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Integration, Indefinite

Free practice problems, each with a complete solution, for typical beginning indefinite integrals in Calculus.

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

Summary: Indefinite Integration
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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
\[\int \! x^n\, dx = \frac{1}{n+1} x^{n+1} + C \qquad (n \ne -1) \]


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 $e^x$
\[\int \! e^x \, dx = e^x+ C \]


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

\[ \int \! k^x = \frac{1}{\ln k} k^x + C \]


Trigonometric Rules: Integrals of Trig Functions

\[ \begin{align*}
\int \cos x \, dx &= \sin x + C \\[8px]
\int \sin x \, dx &= -\cos x + C \\[8px]
\int \sec^2 x \, dx &= \tan x + C \\[8px]
\int \sec x \tan x \, dx &= \sec x + C \\[8px]
\int \csc^2 x \, dx &= -\cot x + C \\[8px]
\end{align*} \]

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

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You can of course practice typical problems below, each with a complete solution.

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. Finally add C." We can express this process mathematically as \[\bbox[yellow,5px]{ \int \! x^n\, dx = \left(\frac{1}{n+1} \right)x^{n+1} + C \qquad (n \ne -1) }\] For example, \[ \begin{align*} \int \! x^3\, dx &= \left(\frac{1}{3+1} \right)x^{3+1} + C \qquad \phantom{(n \ne -1) } \\[8px] &= \frac{1}{4}x^{4} + C \\[8px] \end{align*} \]
Power Rule Problem #1
(a) Find $\displaystyle{\int \!7\, dx.}$
(b) Find: $\displaystyle{\int \! x \, dx.}$
(c) Find : $\displaystyle{\int \! \left(7 + x \right) \, dx }.$
Show/Hide Solution
Solution SummarySolution (a) DetailSolution (b) DetailSolution (c) Detail
(a) $7 x + C$
(b) $\frac{1}{2} x^2 + C$
(c) $7x + \frac{1}{2} x^2 + C$
With a little practice, you will soon remember that the integral $$\int \!dx = x + C$$
Open to see one explanation for 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 \!dx &= \int \! x^0 \, dx \\[8px] &= \left(\frac{1}{0+1}\right)x^{0+1} + C \\[8px] &= x + C \\[8px] \end{align*}
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In this problem, then \[ \begin{align*} \int \!7\, dx &= 7\int dx \\[8px] &= 7 x + C \quad \cmark \end{align*} \]
FAQ: Why isn’t the answer $7(x+C) = 7x + 7C$?
Answer: Remember that the “$+C$” means “plus some constant C,” where we’re not assigning any particular value to C. It doesn’t really make sense, then, to write $+7C$ as the constant-part of the result, since $7C$ is also just some unspecified constant. (It’s just seven times some other unspecified constant.) So traditionally everyone always writes $+C$ to indicate there’s some constant added. You would just never write “$+7C,$” or “$+93C,$” or “$-C,$” or anything else. Always just: “$+C.$”
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. Finally add C.” We can express that process as $$\int \! x^n\, dx = \left( \frac{1}{n+1}\right)x^{n+1} + C \qquad (n \ne -1)$$ In this case, then we have $n = 1$ and so \[ \begin{align*} \int \! x \, dx &= \left( \frac{1}{1+1}\right)x^{1+1} + C \qquad \phantom{(n \ne -1)} \\[8px] &= \frac{1}{2} x^2 + C \quad \cmark \end{align*} \]
We’ve already evaulated the two pieces of this integral in parts (a) and (b), so we can make use of those results: \[ \begin{align*} \int \! \left(7 + x \right) \, dx &= \int \! 7\,dx + \int\!x \,dx \\[8px] &= 7x + \frac{1}{2} x^2 + C \quad \cmark \end{align*} \]
FAQ: Shouldn’t that be $+C_1 + C_2,$ or $+2C,$ or something?
Answer: Remember that the “$+C$” means “plus some constant C,” where we’re not assigning any particular value to C. It doesn’t really make sense, then, to write $+C_1 + C_2$ as the constant-part of the result, since the sum of two unspecified constants is also just some unspecified constant. So traditionally everyone always writes $+C$ to indicate there’s some constant added. You would just never write “$+C_1 + C_2$” or anything else. Always just: “$+C.$”
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Power Rule Problem #2
(a) Find: $\displaystyle{\int\!x^2 \, dx.}$
(b) Find: $\displaystyle{\int \!5x^4 \, dx.}$
(c) Find: $\displaystyle{\int \! \left(x^2 - 5x^4 \right)dx.}$
Show/Hide Solution
Solution SummarySolution (a) DetailSolution (b) DetailSolution (c) Detail
(a) $\dfrac{1}{3} x^{3} +C$
(b) $x^5 + C$
(c) $\dfrac{1}{3}x^3 – x^5 + C$
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. Finally add C.” We can express that process as $$\int \! x^n\, dx = \left(\frac{1}{n+1} \right) x^{n+1}+ C \qquad (n \ne -1)$$ In this case, then \[ \begin{align*} \int\!x^2 \, dx &= \left( \frac{1}{2+1}\right)x^{2+1} +C\qquad \phantom{(n \ne -1) } \\[8px] &= \frac{1}{3} x^{3} +C \quad \cmark \end{align*} \]
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. Finally add C.” We can express that process as $$\int \! x^n\, dx = \left(\frac{1}{n+1} \right) x^{n+1}+ C \qquad (n \ne -1)$$ In this case, then \[ \begin{align*} \int \!5x^4 \, dx &= 5\int_0^2 \!x^4 \, dx \\[8px] &= 5 \left( \frac{1}{4+1}\right)x^{4+1} +C \\[8px] &= 5 \cdot \frac{1}{5}x^5 +C \\[8px] &= x^5 + C \quad \cmark \end{align*} \]
We’ve already evaluated the two pieces of this integral in parts (a) and (b), so we can make use of those results: \[ \begin{align*} \int \! \left(x^2 – 5x^4 \right) dx &= \int \! x^2\, dx – \int \! 5x^4 \, dx \\[8px] &=\dfrac{1}{3}x^3 – x^5 + C \quad \cmark \end{align*} \]
FAQ: Shouldn’t that be $+C_1 + C_2,$ or $+C_1 – C_2,$ or something?
Answer: Remember that the “$+C$” means “plus some constant C,” where we’re not assigning any particular value to C. It doesn’t really make sense, then, to write $+C_1 + C_2$ as the constant-part of the result, since the sum of two unspecified constants is also just some unspecified constant. So traditionally everyone always writes $+C$ to indicate there’s some constant added. You would just never write “$+C_1 + C_2$” or anything else. Always just: “$+C.$”
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Power Rule Problem #3
Evaluate the integral: $\displaystyle{\int \! \left(\frac{2}{3}w^5 - \frac{1}{4}w^3 + w \right) dw.}$
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. Finally add C.” We can express that process as $$\int \! w^n\, dw = \left(\frac{1}{n+1} \right) w^{n+1}+ C \qquad (n \ne -1)$$ 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: \[ \begin{align*} \int \! \left(\frac{2}{3}w^5 – \frac{1}{4}w^3 + w \right) dw &= \frac{2}{3}\int \!w^5 \, dw – \frac{1}{4}\int \!w^3 \, dw + \int \!w \, dw \\[8px] &= \frac{2}{3} \cdot \frac{1}{5+1}x^{5+1} – \frac{1}{4} \cdot \frac{1}{3+1}w^{3+1} + \frac{1}{1+1}w^{1+1} + C\\[8px] &= \frac{2}{3} \cdot \frac{1}{6} x^6 – \frac{1}{4} \cdot \frac{1}{4}w^4 + \frac{1}{2}w^2 + C \\[8px] &= \frac{1}{9} x^6 – \frac{1}{16}x^4 + \frac{1}{2}x^2 +C \quad \cmark \end{align*} \]
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Power Rule Problem #4
Evaluate the integral: $\displaystyle{\int \! \left( x^2 -1\right)^2 \, dx . }$
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: \[ \begin{align*} \int_ \! \left( x^2 -1\right)^2 \, dx &= \int \! \left(x^4 -2x^2 + 1 \right) \, dx \\[8px] &= \int \! x^4 \, dx -2 \int \! x^2 \, dx + \int \! \, dx \\[8px] &= \frac{1}{5}x^5 – 2 \left( \frac{1}{3}x^3 \right) + x +C \\[8px] &= \frac{1}{5}x^5 – \frac{2}{3}x^3 + x +C\quad \cmark \end{align*} \]
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Power Rule Problem #5
Evaluate the integral: $\displaystyle{ \int \!\frac{4x - 7x^5}{x^3} \, dx. }$
Show/Hide Solution
\[ \begin{align*} \int \!\frac{4x – 7x^5}{x^3} \, dx &= \int \!\left( 4x^{-2} – 7x^2\right) \, dx \\[8px] &= 4\int \! x^{-2} \, dx – 7\int \! x^2 \, dx \\[8px] &= 4 \cdot \frac{1}{-2 +1} x^{-2+1} – 7 \cdot \frac{1}{2+1} x^{2+1}+C \\[8px] &= 4 (-1)x^{-1} – 7 \cdot \frac{1}{3} x^3 + C \\[8px] &= \dfrac{-4}{x} – \frac{7}{3} x^3 + C \quad \cmark \end{align*} \]
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Power Rule Problem #6
(a) Evaluate the integral: $\displaystyle{\int\!\sqrt{x} \, dx.}$
(b) Evaluate the integral: $\displaystyle{\int\!\frac{1}{\sqrt{x}} \, dx.}$
(c) Evaluate the integral: $\displaystyle{\int\! \left(\sqrt{x} - \frac{1}{\sqrt{x}} \right)\, dx.}$
Show/Hide Solution
Solution SummarySolution (a) DetailSolution (b) DetailSolution (c) Detail
(a) $\dfrac{2}{3}x^{3/2}+C$
(b) $2 \sqrt{x} + C$
(c) $\dfrac{2}{3}x^{3/2} – 2 \sqrt{x} + C$
\[ \begin{align*} \int\!\sqrt{x} \, dx &= \int\!x^{1/2} \, dx \\[8px] &= \frac{1}{\frac{1}{2}+1} x^{\frac{1}{2} + 1} + C \\[8px] &= \frac{1}{\frac{3}{2}}x^{3/2}+C \\[8px] &= \frac{2}{3}x^{3/2}+C \quad \cmark \end{align*} \]
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. Finally add C.” $$\int \! x^n\, dx = \left(\frac{1}{n+1} \right) x^{n+1} + C \qquad (n \ne -1)$$ In this case, for instance \[ \begin{align*} \int\!\frac{1}{\sqrt{x}} \, dx &= \int\!x^{-1/2} \, dx \\[8px] &= \frac{1}{-\frac{1}{2}+1}x^{-1/2 + 1} + C \\[8px] &= \frac{1}{\frac{1}{2}} x^{1/2} +C \\[8px] &= 2 x^{1/2} +C \quad \cmark \\[8px] &= 2 \sqrt{x} + C \quad \cmark \end{align*} \] Note: The last two lines are equivalent, and either is correct as the final answer.
\[ \begin{align*} \int\! \left(\sqrt{x} – \frac{1}{\sqrt{x}} \right)\, dx &= \int\!x^{1/2} \, dx – \int\!x^{-1/2} \, dx \\[8px] &= \frac{1}{\frac{3}{2}}x^{3/2} – \frac{1}{\frac{1}{2}} x^{1/2} +C \\[8px] &= \frac{2}{3}x^{3/2} – 2 x^{1/2} + C \quad \cmark \\[8px] &= \frac{2}{3}x^{3/2} – 2 \sqrt{x} + C \quad \cmark \end{align*} \]
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Power Rule Problem #7
Find: $\displaystyle{\int \! \frac{x -5}{\sqrt{x}} \, dx.}$
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. Finally add C.” $$\int \! x^n\, dx = \left(\frac{1}{n+1} \right) x^{n+1} + C \qquad (n \ne -1)$$ In this case, then: \[ \begin{align*} \int \! \frac{x -5}{\sqrt{x}} \, dx &= \int \! \left(x^{1/2} – 5 x^{-1/2} \right) \, dx \\[8px] &= \int \! x^{1/2} \, dx – 5 \int \! x^{-1/2}\, dx \\[8px] &= \frac{1}{\frac{3}{2}} x^{3/2} – 5 \cdot \frac{1}{\frac{1}{2}} x^{1/2} + C \\[8px] &= \frac{2}{3}x^{3/2} – 10 x^{1/2} + C \quad \cmark \end{align*} \]
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Power Rule Problem #8
Find $\displaystyle{\int \!\left( 5 \sqrt{x^5} + 2 \sqrt[3]{x^2} \right) \, dx .}$
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. Finally add C.” $$\int \! x^n\, dx = \left(\frac{1}{n+1} \right) x^{n+1} + C \qquad (n \ne -1)$$ In this case then: \[ \begin{align*} \int \!\left( 5 \sqrt{x^5} + 2 \sqrt[3]{x^2} \right) \, dx &= 5\int \! x^{5/2}\, dx + 2\int \! x^{2/3} \, dx \\[8px] &= 5 \cdot \frac{1}{\frac{7}{2}}x^{7/2} + 2 \cdot \frac{1}{\frac{5}{3}} x^{5/3} + C \\[8px] &= 5 \cdot \frac{2}{7}x^{7/2} + 2 \cdot \frac{3}{5}x^{5/3} + C\\[8px] &= \frac{10}{7}x^{7/2} + \frac{6}{5}x^{5/3} + C \quad \cmark \end{align*} \]
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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 $e^x:$ \[ \bbox[yellow,5px]{ \int \! e^x \, dx = e^x + C } \]
Exponential Rule e^x Problem #1
Evaluate the integral: $\displaystyle{\int \! 5e^x \, dx .}$
Show/Hide Solution
\[ \begin{align*} \int \! 5e^x \, dx &= 5 \int \! e^x \, dx \\[8px] &= 5 e^x + C \quad \cmark \end{align*} \]
FAQ: Why isn’t the answer $5(e^x+C) = 5e^x + 5C$?
Answer: Remember that the “$+C$” means “plus some constant C,” where we’re not assigning any particular value to C. It doesn’t really make sense, then, to write $+5C$ as the constant-part of the result, since $5C$ is also just some unspecified constant. (It’s just five times some other unspecified constant.) So traditionally everyone always writes $+C$ to indicate there’s some constant added. You would just never write “$+5C,$” or “$+93C,$” or “$-C,$” or anything else. Always just: “$+C.$”
[hide solution]
Exponential Rule e^x Problem #2
Evaluate the integral: $\displaystyle{\int \! \left( \frac{1}{\sqrt{x^5}} - e^x\right) \, dx .}$
Show/Hide Solution
\[ \begin{align*} \int \! \left( \frac{1}{\sqrt{x^5}} – e^x\right) \, dx &= \int \!x^{-5/2} dx – \int \!e^x \, dx \\[8px] &= \frac{1}{-\frac{5}{2}+1}x^{-5/2 + 1} – e^x + C \\[8px] &= -\frac{2}{3}x^{-3/2} -e^x + C\quad \cmark \end{align*} \]
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III. Exponential Rule: k^x
\[\bbox[yellow,5px]{ \int \! k^x = \frac{1}{\ln k} k^x + C} \]
Open to see how this rule encompasses the e^x rule
Recall that $\ln e = 1$. Then \begin{align*} \int \! e^x &= \frac{1}{\ln e} e^x + C \\[8px] &= e^x + C \end{align*}
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Exponential Rule c^x Problem #1
Find: $\displaystyle{\int \! 3^x \, dx .}$
Show/Hide Solution
\[ \int \! 3^x \, dx = \frac{1}{\ln 3} 3^x + C \quad \cmark \]
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Exponential Rule c^x Problem #2
Find: $\displaystyle{\int \! \left( x^5 + 5^x\right) \, dx }.$
Show/Hide Solution
CAREFUL: This integral has both $x^{\text{(some power)}}$ and $\text{(constant)}^x$. \[ \begin{align*} \int \! \left( x^5 + 5^x\right) \, dx &= \int \! x^5 \, dx + \int \! 5^x\, dx \\[8px] &= \frac{1}{6}x^6 + \frac{1}{\ln 5} 5^x + C \quad \cmark \end{align*} \]
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IV. Trigonometric Rules
We're listing here on the trig integrals that you should know at this early stage because each follows directly from a derivative you know. For example, since $$(\sin x)' = \cos x$$ we know immediately that $$\int \! \cos x \, dx = \sin x + C $$

Accordingly: \[ \bbox[yellow,5px]{\begin{align*} \int \cos x \, dx &= \sin x + C \\[8px] \int \sin x \, dx &= -\cos x + C \\[8px] \int \sec^2 x \, dx &= \tan x + C \\[8px] \int \sec x \tan x \, dx &= \sec x + C \\[8px] \int \csc^2 x \, dx &= -\cot x + C \\[8px] \end{align*}} \] 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: \[ \bbox[10px,border:2px solid blue]{ \begin{align*} \sin^2 x &= \frac{1}{2} - \frac{1}{2}\cos 2x \\ \\ \cos^2 x &= \frac{1}{2} + \frac{1}{2}\cos 2x \end{align*}}\] Then \begin{align*} \int \sin^2 x \, dx &= \int \left(\frac{1}{2} - \frac{1}{2}\cos 2x \right) \, dx = \frac{1}{2}x - \frac{1}{4}\sin 2x + C \quad [*] \\[8px] \int \cos^2 x \, dx &= \int \left( \frac{1}{2} + \frac{1}{2}\cos 2x \right) \, dx = \frac{1}{2}x + \frac{1}{4}\sin 2x + C \quad [*] \\[8px] \end{align*}
Trig Problem #1
Find: $\displaystyle{ \int \!4 \cos x \, dx .}$
Show/Hide Solution
Recall that $\int\cos x \, dx = \sin x + C.$ Then \[ \begin{align*} \int \!4 \cos x \, dx &= 4\int \! \cos x \, dx \\[8px] &= 4 \sin x + C \quad \cmark \end{align*} \]
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Trig Problem #2
Find: $\displaystyle{\int \!\frac{1}{2}\, \sin \theta \, d\theta .}$
Show/Hide Solution
Recall that $\int \sin x \, dx = -\cos x + C.$ Then \[ \begin{align*} \int \! \frac{1}{2}\,\sin \theta \, d\theta &= \frac{1}{2} \int \! \sin \theta \, d\theta \\[8px] &= -\frac{1}{2} \cos \theta + C \quad \cmark \end{align*} \]
[hide solution]
Trig Problem #3
Find: $\displaystyle{\int \frac{dx}{\sec x} .}$
Show/Hide Solution
Don’t forget your basic trig definitions! Here, $\sec x = \dfrac{1}{\cos x}$: \[ \begin{align*} \int\frac{dx}{\sec x} &= \int \frac{dx}{\frac{1}{\cos x}} \\[8px] &= \int \! \cos x \, dx \\[8px] &= \sin x + C \quad \cmark \end{align*} \]
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Trig Problem #4
Find: $\displaystyle{\int \! \sec^2 x \, dx .}$
Show/Hide Solution
Recall that $\int \sec^2 x \, dx = \tan x + C.$ Then \[ \int \! \sec^2 x \, dx = \tan x + C \quad \cmark \]
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Trig Problem #5
Find: $\displaystyle{ \int \! \sec x \tan x \, dx. }$
Show/Hide Solution
Recall that $\int \sec x \tan x \, dx = \sec x + C.$ Then \[ \int \! \sec x \tan x \, dx = \sec x + C \quad \cmark\]
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Trig Problem #6
Find: $\displaystyle{\int \! \left(\csc^2 x + 2x^2 \right) \, dx . }$
Show/Hide Solution
Recall that $\int \csc^2 x \, dx = -\cot x + C.$ Then \[ \begin{align*} \int \! \left(\csc^2 x + 2x^2 \right) \, dx &= \int \! \csc^2 x\, dx + 2\int \! x^2 \, dx \\[8px] &= -\cot x + 2 \cdot \frac{1}{3}x^3+ C \\[8px] &= -\cot x + \frac{2}{3}x^3+ C \quad \cmark \end{align*} \]
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Trig Problem #7
Find: $\displaystyle{\int \! \frac{3 + 2\cos^2 x}{\cos^2 x} \, dx . }$
Show/Hide Solution
\[ \begin{align*} \int \! \frac{3 + 2\cos^2 x}{\cos^2 x} \, dx &= \int\! \frac{3}{\cos^2 x}\, dx \, + \, \int \! \frac{2\cos^2 x}{\cos^2 x} \, dx \\[8px] &= 3\int\! \sec^2 x \, dx \, + \, 2\int \! dx \\[8px] &= 3 \tan x + \, 2 x + C \quad \cmark \end{align*} \]
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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
Find: $\displaystyle{ \int \! \tan^2 x \, dx .}$
Show/Hide Solution
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 $$\tan^2 x = \sec^2 x – 1$$ \[ \begin{align*} \int\! \tan^2 x \, dx &= \int \!\left(\sec^2 x – 1 \right) \, dx \\[8px] &= \int \!\sec^2 x \, dx – \int\! \, dx \\[8px] &= \tan x – x + C \quad \cmark \end{align*} \]
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Trig Problem #9
Find: $\displaystyle{\int \!\frac{5 \sin x + 5 \sin x \tan^2 x}{\sec^2 x} \, dx. }$
Show/Hide Solution
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: \[ \begin{align*} \int \!\frac{5 \sin x + 5 \sin x \tan^2 x}{\sec^2 x} \, dx &= \int \!\frac{5 \sin x \,(1 + \tan^2 x)}{\sec^2 x} \, dx \\[8px] \end{align*} \] Ah, remember the identity $1 + \tan^2 x = \sec^2 x$ \[ \begin{align*} \phantom{ \int \!\frac{5 \sin x + 5 \sin x \tan^2 x}{\sec^2 x} \, dx} &= \int \!\frac{5 \sin x \,(\sec^2 x)}{\sec^2 x} \, dx \\[8px] &= 5 \int \! \sin x \, dx \\[8px] &= -5 \cos x + C \quad \cmark \end{align*} \]
[hide solution]
Trig Problem #10
Evaluate the integral: $\displaystyle{\int \! \frac{\sec^3 x - \sec x}{\tan x} \, dx .}$
Show/Hide Solution
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: \[ \begin{align*} \int \! \frac{\sec^3 x – \sec x}{\tan x} \, dx &= \int \! \frac{\sec x\,(\sec^2 x – 1)}{\tan x} \, dx \end{align*} \] Ah, remember the identity $1 + \tan^2 x = \sec^2 x$, so $\sec^2 x – 1 = \tan^2 x$: \[ \begin{align*} \phantom{\int \! \frac{\sec^3 x – \sec x}{\tan x} \, dx } &= \int \! \frac{\sec x\,(\tan^2 x)}{\tan x}\, dx \\[8px] &= \int \! \sec x \tan x\, dx \\[8px] &= \sec x + C \quad \cmark \end{align*} \]
[hide solution]

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