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Integration, Definite

Free practice problems, each with a complete solution, of computing typical beginning definite integrals that use the power rule, exponentials, and basic trig functions.

You can access our Handy Table of Integrals from the Reference menu at the top of the screen at any time.

Summary: Definite Integration
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Important Notation

When we write $\big[F(x)\big]_a^b$, it means compute $F(b)$ and then subtract $F(a)$:
$$\big[F(x)\big]_a^b = F(b) – F(a)$$

Power Rule: Integration of $x^n$ $(n \ne -1)$

If you’re integrating x-to-some-power (except $x^{-1}$), the rule to remember is: “Increase the power by 1, and then divide by the new power.” We can express this process mathematically as
\[\begin{align*}
\int_a^b \! x^n\, dx &= \frac{1}{n+1}\big[ x^{n+1}\big]_a^b \qquad (n \ne -1) \\[8px]
&= \frac{1}{n+1}\left[b^{n+1} – a^{n+1} \right]
\end{align*}\]


Exponential Rule: Integration of $e^x$

This integral is the easiest to remember: since $(e^x)’ = e^x$, the integral of $e^x$ is also $e^x$
\[ \begin{align*}
\int_a^b \! e^x \, dx &= \big[ e^x\big]_a^b \\[8px]
&= e^b – e^a
\end{align*} \]


Exponential Rule: Integration of $c^x$ (where $c$ is a constant)

\[\begin{align*}
\int_a^b \! c^x &= \frac{1}{\ln c} \big[ c^x \big]_a^b \\[8px]
&= \frac{1}{\ln c} \big[c^b – c^a \big]
\end{align*}\]


Trigonometric Rules: Integrals of Trig Functions

\[ \begin{align*}
\int_a^b \cos x \, dx &= \big[\sin x\big]_a^b \\[8px]
\int_a^b \sin x \, dx &= -\big[\cos x\big]_a^b \\[8px]
\int_a^b \sec^2 x \, dx &= \big[\tan x\big]_a^b \\[8px]
\int_a^b \sec x \tan x \, dx &= \big[\sec x\big]_a^b \\[8px]
\int_a^b \csc^2 x \, dx &= -\big[\cot x\big]_a^b \\[8px]
\end{align*} \]

You might also be expected to know the integrals for $\sin^2 x$ and $\cos^2 x$, because they follow immediately from use of the half-angle formulas:

\begin{align*}
\int_a^b \sin^2 x \, dx &= \int_a^b \left(\frac{1}{2} – \frac{1}{2}\cos 2x \right) \, dx = \big[\frac{1}{2}x – \frac{1}{4}\sin 2x \big]_a^b \\[8px]
\int_a^b \cos^2 x \, dx &= \int_a^b \left( \frac{1}{2} + \frac{1}{2}\cos 2x \right) \, dx = \big[\frac{1}{2}x + \frac{1}{4}\sin 2x \big]_a^b
\end{align*}

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You can practice each of these common integration problems below.

I. Power Rule
If you're integrating x-to-some-power (except $x^{-1}$), the rule to remember is: "Increase the power by 1, and then divide by the new power." We can express this process mathematically as \[\bbox[yellow,5px]{ \begin{align*} \int_a^b \! x^n\, dx &= \frac{1}{n+1}\big[ x^{n+1}\big]_a^b \qquad (n \ne -1) \\[8px] &= \frac{1}{n+1}\left[b^{n+1} - a^{n+1} \right] \end{align*}} \] For example, \[ \begin{align*} \int_a^b \! x^3\, dx &= \frac{1}{3+1}\big[ x^{3+1}\big]_a^b \qquad \phantom{(n \ne -1) } \\[8px] &= \frac{1}{4}\big[ x^{4}\big]_a^b \\[8px] &= \frac{1}{4} \left[b^4 - a^4 \right] \end{align*} \]
Power Rule Problem #1
(a) Evaluate the integral: $\displaystyle{\int_1^5 \!7\, dx.}$
(b) Evaluate the integral: $\displaystyle{\int_1^5 \! x \, dx.}$
(c) Evaluate the integral: $\displaystyle{\int_1^5 \! \left(7 + x \right) \, dx }.$
Show/Hide Solution
Solution SummarySolution (a) DetailSolution (b) DetailSolution (c) Detail
(a) 28
(b) 12
(c) 40
With a little practice, you will soon remember that the integral $$\int_a^b \!dx = \big[x\big]_a^b = b – a$$
Open to see one explanation of this result
Let’s make use of the fact that $1 = x^0$, and use our usual rule for integrating x-to-some-power: \begin{align*} \int_a^b \!dx &= \int_a^b \! x^0 \, dx \\[8px] &= \frac{1}{0+1}\left[x^{0+1} \right]_a^b \\[8px] &= \big[x\big]_a^b \\[8px] &= (b – a) \end{align*}
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In this problem, then \[ \begin{align*} \int_1^5 \!7\, dx &= 7\int_1^5 dx \\[8px] &= 7 \big[x \big]_1^5 \\[8px] &= 7 \cdot (5 – 1) \\[8px] &= 7 \cdot4 = 28 \quad \cmark \end{align*} \]
When you’re integrating x-to-some-power, the rule to remember is: “Increase the power by 1, and then divide by the new power.” We can express that process as $$\int_a^b \! x^n\, dx = \frac{1}{n+1}\left[ x^{n+1}\right]_a^b \qquad (n \ne -1)$$ In this case, then we have $n = 1$ and so \[ \begin{align*} \int_1^5 \! x \, dx &= \frac{1}{1+1}\left[x^{1+1} \right]_1^5 \qquad \phantom{(n \ne -1)} \\[8px] &= \frac{1}{2} \left[x^2 \right]_1^5 \\[8px] &= \frac{1}{2}\left[ 5^2 – 1^2\right] \\[8px] &= \frac{1}{2}\left[25 – 1 \right] \\[8px] &= \frac{1}{2}\left[24 \right] = 12 \quad \cmark \end{align*} \]
We’ve already evaulated the two pieces of this integral in parts (a) and (b), so we can make use of those results: \[ \begin{align*} \int_1^5 \! \left(7 + x \right) \, dx &= \int_1^5 \! 7\,dx + \int_1^5\!x \,dx \\[8px] &= 28 + 12 \\[8px] &= 40 \quad \cmark \end{align*} \] If you had gotten this question without having done parts (a) and (b) first, your solution would probably look something like this: \[ \begin{align*} \int_1^5 \! \left(7 + x \right) \, dx &= \int_1^5 \! 7\,dx + \int_1^5\!x \,dx \\[8px] &= 7 \big[x\big]_1^5 + \frac{1}{2}\left[x^2 \right]_1^5 \\[8px] &= 7 \cdot (5 -1) + \frac{1}{2}(25 – 1) \\[8px] &= 28 + 12 \\[8px] &= 40 \quad \cmark \end{align*} \]
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Power Rule Problem #2
(a) Evaluate the integral: $\displaystyle{\int_0^2\!x^2 \, dx.}$
(b) Evaluate the integral: $\displaystyle{\int_0^2 \!5x^4 \, dx.}$
(c) Evaluate the integral: $\displaystyle{\int_0^2 \! \left(x^2 - 5x^4 \right)dx.}$
Show/Hide Solution
Solution SummarySolution (a) DetailSolution (b) DetailSolution (c) Detail
(a) $ \dfrac{8}{3}$
(b) 32
(c) $-29\dfrac{1}{3}$
When you’re integrating x-to-some-power, the rule to remember is: “Increase the power by 1, and then divide by the new power.” We can express that process as $$\int_a^b \! x^n\, dx = \frac{1}{n+1}\left[ x^{n+1}\right]_a^b \qquad (n \ne -1)$$ In this case, then \[ \begin{align*} \int_0^2\!x^2 \, dx &= \frac{1}{2+1}\left[x^{2+1} \right]_0^2 \qquad \phantom{(n \ne -1) } \\[8px] &= \frac{1}{3} \left[x^{3} \right]_0^2 \\[8px] &= \frac{1}{3}\left[2^3 – 0 \right] = \frac{1}{3}\left(8 \right) \\[8px] &= \frac{8}{3} \quad \cmark \end{align*} \]
When you’re integrating x-to-some-power, the rule to remember is: “Increase the power by 1, and then divide by the new power.” We can express that process as $$\int_a^b \! x^n\, dx = \frac{1}{n+1}\left[ x^{n+1}\right]_a^b \qquad (n \ne -1)$$ In this case, then \[ \begin{align*} \int_0^2 \!5x^4 \, dx &= 5\int_0^2 \!x^4 \, dx \\[8px] &= 5 \cdot \frac{1}{4+1}\left[x^{4+1} \right]_0^2 \\[8px] &= 5 \cdot \frac{1}{5}\left[x^5 \right]_0^2 = \left[x^5 \right]_0^2 \\[8px] &= \left[2^5 – 0 \right] = 32 \quad \cmark \end{align*} \]
We’ve already evaulated the two pieces of this integral in parts (a) and (b), so we can make use of those results: \[ \begin{align*} \int_0^2 \! \left(x^2 – 5x^4 \right) dx &= \int_0^2 \! x^2\, dx – \int_0^2 \! 5x^4 \, dx \\[8px] &= \frac{8}{3} – 32 \\[8px] &= -29\frac{1}{3} \end{align*} \] If you had gotten this question without having done parts (a) and (b) first, your solution would probably look something like this: \[ \begin{align*} \int_0^2 \! \left(x^2 – 5x^4 \right) dx &= \int_0^2 \! x^2\, dx – \int_0^2 \! 5x^4 \, dx \\[8px] &= \frac{1}{3}\left[x^3 \right]_0^2 – 5 \cdot \frac{1}{5}\left[ x^5\right]_0^2 \\[8px] &= \frac{1}{3}\left[2^3 – 0 \right] -[2^5 – 0] \\[8px] &= \frac{8}{3} – 32 \\[8px] &= -29\frac{1}{3} \quad \cmark \end{align*} \]
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Power Rule Problem #3
Evaluate the integral: $\displaystyle{\int_{-3}^0 \! \left(\frac{2}{3}w^5 - \frac{1}{4}w^3 + w \right) dw.}$
Show/Hide Solution
When you’re integrating x-to-some-power, the rule to remember is: “Increase the power by 1, and then divide by the new power.” We can express that process as $$\int_a^b \! w^n\, dw = \frac{1}{n+1}\left[ w^{n+1}\right]_a^b \qquad (n \ne -1)$$ We’ll write out some extra steps to show more detail than usual since this is probably new to you. Once you get the hang of these, you won’t write out every line here: \[ \begin{align*} \int_{-3}^0 \! \left(\frac{2}{3}w^5 – \frac{1}{4}w^3 + w \right) dw &= \frac{2}{3}\int_{-3}^0 \!w^5 \, dw – \frac{1}{4}\int_{-3}^0 \!w^3 \, dw + \int_{-3}^0 \!w \, dw \\[8px] &= \frac{2}{3} \cdot \frac{1}{5+1}\left[x^{5+1} \right]_{-3}^0 – \frac{1}{4} \cdot \frac{1}{3+1}\left[w^{3+1} \right]_{-3}^0 + \frac{1}{1+1}\left[w^{1+1} \right]_{-3}^0 \\[8px] &= \frac{2}{3} \cdot \frac{1}{6}\left[x^6 \right]_{-3}^0 – \frac{1}{4} \cdot \frac{1}{4}\left[w^4 \right]_{-3}^0 + \frac{1}{2}\left[w^2 \right]_{-3}^0 \\[8px] &= \frac{1}{9} \left[0 – (-3)^6 \right] – \frac{1}{16}\left[0 – (-3)^4 \right] + \frac{1}{2}\left[0 – (-3)^2 \right] \\[8px] &= \frac{1}{9}\left[-729 \right] – \frac{1}{16}\left[ -81 \right] + \frac{1}{2} \left[-9 \right] \\[8px] &= -81 +\frac{81}{16} – \frac{9}{2} \\[8px] &= -\frac{1296}{16} + \frac{81}{16} – \frac{72}{16}\\[8px] &= -\frac{1287}{16} \quad \cmark \end{align*} \]
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Power Rule Problem #4
Evaluate the integral: $\displaystyle{\int_5^0 \! \left( x^2 -1\right)^2 \, dx . }$
Show/Hide Solution
Later we’ll learn a technique for dealing with an integral of the form $(…)^n$, but for the time being we must rewrite this integral so that each term is of the form $x^n.$ That means we must expand the quadratic. $$\left( x^2 -1\right)^2 = x^4 -2x^2 + 1$$ We can then use our usual power rule for integrals: \[ \begin{align*} \int_5^0 \! \left( x^2 -1\right)^2 \, dx &= \int_5^0 \! \left(x^4 -2x^2 + 1 \right) \, dx \\[8px] &= \frac{1}{5}\left[x^5 \right]_5^0 – 2 \cdot \frac{1}{3}\left[x^3 \right]_5^0 + \big[x \big]_5^0 \\[8px] &= \frac{1}{5}[0 – 5^5 ] – \frac{2}{3} \left[0 – 5^3 \right] + [0 – 5] \\[8px] &= -5^4 – \frac{2}{3}[-125] -5 \\[8px] &= -625 + \frac{250}{3} – 5 \\[8px] &= -\frac{1640}{3} \quad \cmark \end{align*} \]
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Power Rule Problem #5
Evaluate the integral: $\displaystyle{ \int_1^2 \!\frac{4x - 7x^5}{x^3} \, dx. }$
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\[ \begin{align*} \int_1^2 \!\frac{4x – 7x^5}{x^3} \, dx &= \int_1^2 \!\left( 4x^{-2} – 7x^2\right) \, dx \\[8px] &= 4\int_1^2 \! x^{-2} \, dx – 7\int_1^2 \! x^2 \, dx \\[8px] &= 4 \cdot \frac{1}{-2 +1} \left[x^{-2+1} \right]_1^2 – 7 \cdot \frac{1}{2+1} \left[ x^{2+1}\right]_1^2 \\[8px] &= 4 \cdot (-1)\left[x^{-1} \right]_1^2 – 7 \cdot \frac{1}{3}\left[ x^3\right]_1^2 \\[8px] &= -4 \left[2^{-1} – 1^{-1} \right] – \frac{7}{3} \left[2^3 – 1^3 \right] \\[8px] &= -4 \left[ \frac{1}{2} – 1 \right]- \frac{7}{3} [8 – 1] \\[8px] &= -4 \left[-\frac{1}{2} \right] – \frac{7}{3}[7] \\[8px] &= 2 – \frac{49}{3} \\[8px] &= \frac{6 – 49}{3} \\[8px] &= -\frac{43}{3} \quad \cmark \end{align*} \]
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Power Rule Problem #6
(a) Evaluate the integral: $\displaystyle{\int_1^9\!\sqrt{x} \, dx.}$
(b) Evaluate the integral: $\displaystyle{\int_1^9\!\frac{1}{\sqrt{x}} \, dx.}$
(c) Evaluate the integral: $\displaystyle{\int_1^9\! \left(\sqrt{x} - \frac{1}{\sqrt{x}} \right)\, dx.}$
Show/Hide Solution
Solution SummarySolution (a) DetailSolution (b) DetailSolution (c) Detail
(a) $\dfrac{52}{3}$
(b) 4
(c) $ \dfrac{40}{3}$
We think it’s easiest to work with radicals by rewriting the expression with an exponent, since we can then easily use our usual power rule, “Increase the power by 1, and then divide by the new power.” $$\int_a^b \! x^n\, dx = \frac{1}{n+1}\left[ x^{n+1}\right]_a^b \qquad (n \ne -1)$$ In this case, for instance \[ \begin{align*} \int_1^9\!\sqrt{x} \, dx &= \int_1^9\!x^{1/2} \, dx \\[8px] &= \frac{1}{\frac{1}{2}+1}\left[ x^{\frac{1}{2} + 1} \right]_1^9 \\[8px] &= \frac{1}{\frac{3}{2}}\left[x^{3/2} \right]_1^9 \\[8px] &= \frac{2}{3}\left[9^{3/2} – 1^{3/2} \right] \\[8px] &= \frac{2}{3} \left[27 – 1 \right] \\[8px] &= \frac{2}{3}[26] \\[8px] &= \frac{52}{3} \quad \cmark \end{align*} \]
We think it’s easiest to work with radicals by rewriting the expression with an exponent, since we can then easily use our usual power rule, “Increase the power by 1, and then divide by the new power.” $$\int_a^b \! x^n\, dx = \frac{1}{n+1}\left[ x^{n+1}\right]_a^b \qquad (n \ne -1)$$ In this case, for instance \[ \begin{align*} \int_1^9\!\frac{1}{\sqrt{x}} \, dx &= \int_1^9\!x^{-1/2} \, dx \\[8px] &= \frac{1}{-\frac{1}{2}+1}\left[x^{-1/2 + 1} \right]_1^9 \\[8px] &= \frac{1}{\frac{1}{2}}\left[ x^{1/2}\right]_1^9 \\[8px] &= 2 [9^{1/2} – 1^{1/2}] \\[8px] &= 2[3 – 1]= 2[2] \\[8px] &= 4 \quad \cmark \end{align*} \]
We’ve already evaulated the two pieces of this integral in parts (a) and (b), so we can make use of those results: \[ \begin{align*} \int_1^9\! \left(\sqrt{x} – \frac{1}{\sqrt{x}} \right)\, dx &= \int_1^9\!x^{1/2} \, dx – \int_1^9\!x^{-1/2} \, dx \\[8px] &= \frac{52}{3} – 4 \\[8px] &= \frac{40}{3} \quad \cmark \end{align*} \] If you had gotten this question without having done parts (a) and (b) first, your solution would probably look something like this: \[ \begin{align*} \int_1^9\! \left(\sqrt{x} – \frac{1}{\sqrt{x}} \right)\, dx &= \int_1^9\!x^{1/2} \, dx – \int_1^9\!x^{-1/2} \, dx \\[8px] &= \frac{2}{3}\left[ x^{3/2}\right]_1^9 – 2 \left[ x^{1/2}\right]_1^9 \\[8px] &= \frac{2}{3}[27 -1] – 2[3 – 1] \\[8px] &= \frac{52}{3} – 4 \\[8px] &= \frac{40}{3} \quad \cmark \end{align*} \]
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Power Rule Problem #7
Evaluate the integral: $\displaystyle{\int_4^9 \! \frac{x -5}{\sqrt{x}} \, dx.}$
Show/Hide Solution
\[ \begin{align*} \int_4^9 \! \frac{x -5}{\sqrt{x}} \, dx &= \int_4^9 \! \left(x^{1/2} – 5 x^{-1/2} \right) \, dx \\[8px] &= \frac{1}{\frac{3}{2}} \left[x^{3/2} \right]_4^9 – 5 \cdot \frac{1}{\frac{1}{2}} \left[ x^{1/2} \right]_4^9 \\[8px] &= \frac{2}{3}\left[9^{3/2} – 4^{3/2} \right] – 5 \cdot 2 \left[ 9^{1/2} – 4^{1/2}\right] \\[8px] &= \frac{2}{3}[27 – 8] – 10[3 -2] \\[8px] &= \frac{2}{3}[19] – 10 \\[8px] &= \frac{8}{3} \quad \cmark \end{align*} \]
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Power Rule Problem #8
Evaluate the integral: $\displaystyle{\int_0^1 \!\left( 5 \sqrt{x^5} + 2 \sqrt[3]{x^2} \right) \, dx .}$
Show/Hide Solution
We think it’s easiest to work with radicals by rewriting the expression with an exponent, since we can then easily use our usual power rule, “Increase the power by 1, and then divide by the new power.” $$\int_a^b \! x^n\, dx = \frac{1}{n+1}\left[ x^{n+1}\right]_a^b \qquad (n \ne -1)$$ In this case, for instance \[ \begin{align*} \int_0^1 \!\left( 5 \sqrt{x^5} + 2 \sqrt[3]{x^2} \right) \, dx &= 5\int_0^1 \! x^{5/2}\, dx + 2\int_0^1 \! x^{2/3} \, dx \\[8px] &= 5 \cdot \frac{1}{\frac{7}{2}}\left[ x^{7/2}\right]_0^1 + 2 \cdot \frac{1}{\frac{5}{3}}\left[ x^{5/3}\right]_0^1 \\[8px] &= 5 \cdot \frac{2}{7} [1 – 0] + 2 \cdot \frac{3}{5}[1 – 0] \\[8px] &= \frac{10}{7} + \frac{6}{5} = \frac{50}{35} + \frac{42}{35} \\[8px] &= \frac{92}{35} \quad \cmark \end{align*} \]
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II. Exponential Rule: e^x
This integral is the easiest to remember: since $(e^x)' = e^x$, the integral of $e^x$ is also $e^x$ \[ \bbox[yellow,5px]{ \begin{align*} \int_a^b \! e^x \, dx &= \big[ e^x\big]_a^b \\[8px] &= e^b - e^a \end{align*}} \]
Exponential Rule e^x Problem #1
Evaluate the integral: $\displaystyle{\int_1^3 \! 5e^x \, dx .}$
Show/Hide Solution
\[ \begin{align*} \int_1^3 \! 5e^x \, dx &= 5 \int_1^3 \! e^x \, dx \\[8px] &= 5 \big[ e^x\big]_1^3 \\[8px] &= 5 [e^3 – e] \quad \cmark \end{align*} \]
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Exponential Rule e^x Problem #2
Evaluate the integral: $\displaystyle{\int_1^4 \! \left( \frac{1}{\sqrt{x^5}} - e^x\right) \, dx .}$
Show/Hide Solution
\[ \begin{align*} \int_1^4 \! \left( \frac{1}{\sqrt{x^5}} – e^x\right) \, dx &= \int_1^4 \!x^{-5/2} dx – \int_1^4 \!e^x \, dx \\[8px] &= \frac{1}{-\frac{5}{2}+1}\big[x^{-5/2 + 1} \big]_1^4 – \big[e^x \big]_1^4 \\[8px] &= -\frac{2}{3}\big[x^{-3/2} \big]_1^4 – \big[e^x \big]_1^4 \\[8px] &= -\frac{2}{3} \left[4^{-3/2} – 1 \right] – \left[ e^4 – e\right] \\[8px] &= -\frac{2}{3} \left[\frac{1}{8} – 1 \right] – e^4 + e \\[8px] &= -\frac{2}{3} \left[-\frac{7}{8} \right]- e^4 + e \\[8px] &= \frac{7}{12} – e^4 + e \quad \cmark \end{align*} \]
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III. Exponential Rule: c^x
Here $c$ is a constant (a number, any number). Then \[\bbox[yellow,5px]{ \begin{align*} \int_a^b \! c^x &= \frac{1}{\ln c} \big[ c^x \big]_a^b \\[8px] &= \frac{1}{\ln c} \big[c^b - c^a \big] \end{align*} } \]
Open to see how this rule encompasses the e^x rule
Recall that $\ln e = 1$. Then \begin{align*} \int_a^b \! e^x &= \frac{1}{\ln e} \big[ e^x \big]_a^b \\[8px] &= \big[ e^x \big]_a^b = \big[e^b – e^a \big] \end{align*}
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Exponential Rule c^x Problem #1
Evaluate the integral: $\displaystyle{\int_0^2 \! 3^x \, dx .}$
Show/Hide Solution
\[ \begin{align*} \int_0^2 \! 3^x \, dx &= \frac{1}{\ln 3} \left[3^x \right]_0^2\\[8px] &= \frac{1}{\ln 3} \left[3^2 – 3^0 \right] \\[8px] &= \frac{1}{\ln 3}[9-1] \\[8px] &= \frac{8}{\ln 3} \quad \cmark \end{align*} \]
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Exponential Rule c^x Problem #2
Evaluate the integral: $\displaystyle{\int_0^1 \! \left( x^5 + 5^x\right) \, dx }.$
Show/Hide Solution
CAREFUL: This integral has both $x^{\text{(some power)}}$ and $\text{(constant)}^x$. \[ \begin{align*} \int_0^1 \! \left( x^5 + 5^x\right) \, dx &= \int_0^1 \! x^5 \, dx + \int_0^1 \! 5^x\, dx \\[8px] &= \frac{1}{6}\big[x^6 \big]_0^1 + \frac{1}{\ln 5} \big[5^x \big]_0^1 \\[8px] &= \frac{1}{6}\big[1^6 – 0^6 \big] + \frac{1}{\ln 5} \big[5^1 – 5^0 \big] \\[8px] &= \frac{1}{6} + \frac{1}{\ln 5}[5 – 1] \\[8px] &= \frac{1}{6} + \frac{4}{\ln 5} \quad \cmark \end{align*} \]
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IV. Trigonometric Rules
We're listing here only the trig integrals that you should be familiar with at this early stage: each of these follows directly from a derivative you immediately know. For example, since $$(\sin x)' = \cos x$$ we know immediately that $$\int_a^b \! \cos x \, dx = \big[ \sin x \big]_a^b$$ Accordingly: \[ \bbox[yellow,5px]{\begin{align*} \int_a^b \cos x \, dx &= \big[\sin x\big]_a^b \\[8px] \int_a^b \sin x \, dx &= -\big[\cos x\big]_a^b \\[8px] \int_a^b \sec^2 x \, dx &= \big[\tan x\big]_a^b \\[8px] \int_a^b \sec x \tan x \, dx &= \big[\sec x\big]_a^b \\[8px] \int_a^b \csc^2 x \, dx &= -\big[\cot x\big]_a^b \\[8px] \end{align*}} \] Each integral follows directly from a derivative you know. You can review those using our Trig Function Derivatives table; it's always available from the Reference menu at the top of every page.
You might also be expected to know the integrals for $\sin^2 x$ and $\cos^2 x$, because they follow immediately from use of the half-angle formulas: \[ \bbox[10px,border:2px solid blue]{ \begin{align*} \sin^2 x &= \frac{1}{2} - \frac{1}{2}\cos 2x \\ \\ \cos^2 x &= \frac{1}{2} + \frac{1}{2}\cos 2x \end{align*}}\] Then \begin{align*} \int_a^b \sin^2 x \, dx &= \int \left(\frac{1}{2} - \frac{1}{2}\cos 2x \right) \, dx = \left[\frac{1}{2}x - \frac{1}{4}\sin 2x\right]_a^b \\[8px] \int_a^b \cos^2 x \, dx &= \int \left( \frac{1}{2} + \frac{1}{2}\cos 2x \right) \, dx = \left[\frac{1}{2}x + \frac{1}{4}\sin 2x\right]_a^b \end{align*}
Trig Problem #1
Evaluate the integral: $\displaystyle{ \int_0^{\pi/2} \!4 \cos x \, dx .}$
Show/Hide Solution
Recall that $\int_a^b \cos x \, dx = \big[\sin x\big]_a^b.$ Then \[ \begin{align*} \int_0^{\pi/2} \!4 \cos x \, dx &= 4\int_0^{\pi/2} \! \cos x \, dx \\[8px] &= 4 \big[\sin x \big]_0^{\pi/2} \\[8px] &= 4 \big[\sin \frac{\pi}{2} – \sin 0 \big] \\[8px] &= 4 \big[1 – 0 \big] \\[8px] &= 4 \quad \cmark \end{align*} \]
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Trig Problem #2
Evaluate the integral: $\displaystyle{\int_0^{\pi} \!\frac{1}{2}\, \sin \theta \, d\theta .}$
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Recall that $\int_a^b \sin x \, dx = -\big[\cos x\big]_a^b.$ Then \[ \begin{align*} \int_0^{\pi} \! \frac{1}{2}\,\sin \theta \, d\theta &= \frac{1}{2} \int_0^{\pi} \! \sin \theta \, d\theta \\[8px] &= -\frac{1}{2} \big[ \cos \theta \big]_0^{\pi} \\[8px] &= -\frac{1}{2} \big[\cos \pi – \cos 0 \big] \\[8px] &= -\frac{1}{2} [-1 – 1] = -\frac{1}{2} [-2] \\[8px] &= 1 \quad \cmark \end{align*} \]
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Trig Problem #3
Evaluate the integral: $\displaystyle{\int_0^{\pi/4} \frac{dx}{\sec x} .}$
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Don’t forget your basic trig definitions, for instance $\sec x = \dfrac{1}{\cos x}$: \[ \begin{align*} \int_0^{\pi/4} \frac{dx}{\sec x} &= \int_0^{\pi/4} \frac{dx}{\frac{1}{\cos x}} \\[8px] &= \int_0^{\pi/4} \! \cos x \, dx \\[8px] &= \big[\sin x \big]_0^{\pi/4} \\[8px] &= \sin \frac{\pi}{4} – \sin 0 \\[8px] &= \frac{\sqrt{2}}{2} – 0 \\[8px] &= \frac{\sqrt{2}}{2} \quad \cmark \end{align*} \]
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Trig Problem #4
Evaluate the integral: $\displaystyle{\int_0^{\pi/4} \! \sec^2 x \, dx .}$
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Recall that $\int_a^b \sec^2 x \, dx = \big[\tan x\big]_a^b.$ Then \[ \begin{align*} \int_0^{\pi/4} \! \sec^2 x \, dx &= \big[\tan x \big]_0^{\pi/4} \\[8px] &= \tan \frac{\pi}{4} – \tan 0 \\[8px] &= 1 – 0 \\[8px] &= 1 \quad \cmark \end{align*} \]
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Trig Problem #5
Evaluate the integral: $\displaystyle{ \int_{\pi/6}^{\pi/3} \! \sec x \tan x \, dx. }$
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Recall that $\int_a^b \sec x \tan x \, dx = \big[\sec x\big]_a^b.$ Then \[ \begin{align*} \int_{\pi/6}^{\pi/3} \! \sec x \tan x \, dx &= \big[\sec x \big]_{\pi/6}^{\pi/3} \\[8px] &= \left[ \sec \frac{\pi}{3} – \sec \frac{\pi}{6}\right] \\[8px] &= 2 – \frac{2}{\sqrt{3}} \quad \cmark \end{align*} \]
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Trig Problem #6
Evaluate the integral: $\displaystyle{\int_{\pi/4}^{3\pi/4} \! \left(\csc^2 x + 2x^2 \right) \, dx . }$
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Recall that $\int_a^b \csc^2 x \, dx = -\big[\csc x\big]_a^b.$ Then \[ \begin{align*} \int_{\pi/4}^{3\pi/4} \! \left(\csc^2 x + 2x^2 \right) \, dx &= \int_{\pi/4}^{3\pi/4} \! \csc^2 x\, dx + 2\int_{\pi/4}^{3\pi/4} \! x^2 \, dx \\[8px] &= -\big[\cot x \big]_{\pi/4}^{3\pi/4} + 2 \cdot \frac{1}{3} \left[ x^3\right]_{\pi/4}^{3\pi/4} \\[8px] &= – \big[\cot \frac{3\pi}{4} – \cot \frac{\pi}{4} \big] + \frac{2}{3} \left[ \left(\frac{3\pi}{4} \right)^3 – \left( \frac{\pi}{4} \right)^3\right] \\[8px] &= -[-1 – 1] + \left[\frac{27 \pi^3}{64} – \frac{\pi^3}{64} \right] \\[8px] &=2 + \frac{2}{3}\left[ \frac{26\pi^3}{64}\right] \\[8px] &= 2 + \frac{13 \pi^3}{48} \quad \cmark \end{align*} \]
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Trig Problem #7
Evaluate the integral: $\displaystyle{\int_0^{\pi/3} \! \frac{3 + 2\cos^2 x}{\cos^2 x} \, dx . }$
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\[ \begin{align*} \int_0^{\pi/3} \! \frac{3 + 2\cos^2 x}{\cos^2 x} \, dx &= \int_0^{\pi/3} \! \frac{3}{\cos^2 x}\, dx \, + \, \int_0^{\pi/3} \! \frac{2\cos^2 x}{\cos^2 x} \, dx \\[8px] &= 3\int_0^{\pi/3} \! \sec^2 x \, dx \, + \, 2\int_0^{\pi/3} \! dx \\[8px] &= 3 \big[ \tan x\big]_0^{\pi/3} + \, 2 \big[x \big]_0^{\pi/3} \\[8px] &= 3 \left[ \tan \frac{\pi}{3} – \tan 0\right] + 2 \big[\frac{\pi}{3} – 0 \big] \\[8px] &= 3 \big[\sqrt{3} – 0 \big] + \frac{2\pi}{3} \\[8px] &= 3 \sqrt{3} + \frac{2\pi}{3} \quad \cmark \end{align*} \]
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You may also be expected to use the Trig Identity and its variants: \[ \bbox[10px,border:2px solid blue]{ \begin{align*} \sin^2 x + \cos^2 x &= 1 \\[8px] 1 + \cot^2 x &= \csc^2 x \\[8px] 1 + \tan^2 x &= \sec^2 x \end{align*}}\] We'll of course illustrate the use of these identities in the problems below.
Trig Problem #8
Evaluate the integral: $\displaystyle{ \int_0^{\pi/4} \! \tan^2 x \, dx .}$
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When you see $\tan^2 x$, you should think of the trig identity $$1 + \tan^2 x = \sec^2 x$$ We can thus make the substitution $$\tan^2 x = \sec^2 x – 1$$ \[ \begin{align*} \int_0^{\pi/4} \! \tan^2 x \, dx &= \int_0^{\pi/4} \!\left(\sec^2 x – 1 \right) \, dx \\[8px] &= \int_0^{\pi/4} \!\sec^2 x \, dx – \int_0^{\pi/4} \! \, dx \\[8px] &= \big[\tan x \big]_0^{\pi/4} – \big[x \big]_0^{\pi/4} \\[8px] &= \left[\tan \frac{\pi}{4} – \tan 0 \right] – \big[\frac{\pi}{4} – 0 \big] \\[8px] &= \big[1 – 0 \big] – \frac{\pi}{4} \\[8px] &= 1 – \frac{\pi}{4} \quad \cmark \end{align*} \]
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Trig Problem #9
Evaluate the integral: $\displaystyle{\int_0^{\pi/4} \!\frac{5 \sin x + 5 \sin x \tan^2 x}{\sec^2 x} \, dx. }$
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It’s probably not immediately obvious what to do here. In that case, your best bet is to dive in and start by doing whatever you can see to do. We notice that both terms in the numerator have $5\sin x$, so let’s start by factoring that out: \[ \begin{align*} \int_0^{\pi/4} \!\frac{5 \sin x + 5 \sin x \tan^2 x}{\sec^2 x} \, dx &= \int_0^{\pi/4} \!\frac{5 \sin x \,(1 + \tan^2 x)}{\sec^2 x} \, dx \\[8px] \end{align*} \] Ah, remember the identity $1 + \tan^2 x = \sec^2 x$ \[ \begin{align*} \phantom{ \int_0^{\pi/4} \!\frac{5 \sin x + 5 \sin x \tan^2 x}{\sec^2 x} \, dx} &= \int_0^{\pi/4} \!\frac{5 \sin x \,(\sec^2 x)}{\sec^2 x} \, dx \\[8px] &= 5 \int_0^{\pi/4} \! \sin x \, dx \\[8px] &= -5 \big[ \cos x\big]_0^{\pi/4} \\[8px] &= -5 \big[\cos \frac{\pi}{4} – \cos 0 \big] \\[8px] &= -5 \big[ \frac{1}{\sqrt{2}} – 1\big] \\[8px] &= 5 \big[1 – \frac{1}{\sqrt{2}}\big] = \frac{5}{\sqrt{2}}\big[\sqrt{2} – 1 \big] \quad \cmark \end{align*} \]
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Trig Problem #10
Evaluate the integral: $\displaystyle{\int_{\pi/6}^{\pi/3} \! \frac{\sec^3 x - \sec x}{\tan x} \, dx .}$
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It’s probably not immediately obvious how to proceed here. In that case, your best bet is to dive in and start by doing whatever you can see to do. We notice that both terms in the numerator have $\sec x$, so let’s start by factoring that out: \[ \begin{align*} \int_{\pi/6}^{\pi/3} \! \frac{\sec^3 x – \sec x}{\tan x} \, dx &= \int_{\pi/6}^{\pi/3} \! \frac{\sec x\,(\sec^2 x – 1)}{\tan x} \, dx \end{align*} \] Ah, remember the identity $1 + \tan^2 x = \sec^2 x$, so $\sec^2 x – 1 = \tan^2 x$: \[ \begin{align*} \phantom{\int_{\pi/6}^{\pi/3} \! \frac{\sec^3 x – \sec x}{\tan x} \, dx } &= \int_{\pi/6}^{\pi/3} \! \frac{\sec x\,(\tan^2 x)}{\tan x}\, dx \\[8px] &= \int_{\pi/6}^{\pi/3} \! \sec x \tan x\, dx \\[8px] &= \big[\sec x \big]_{\pi/6}^{\pi/3} \\[8px] &= \sec \frac{\pi}{3} – \sec \frac{\pi}{6} \\[8px] &= 2 – \frac{2}{\sqrt{3}} \quad \cmark \end{align*} \]
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