Integrals Involving Inverse Trigonometric Functions

Inverse trigonometric functions appear naturally when integrating expressions of the forms:

• $$\frac{1}{\sqrt{a^{2}-x^{2}}​$$

• $$\frac{1}{a^{2}+x^{2}}​$$

• $$\frac{1}{x\sqrt{x^{2}-a{2}}$$​

These integrals come from reversing derivatives of inverse trig functions.

Derivative Review

Here are the key derivative formulas to remember:

$$\frac{d}{dx}(\sin^{-1}(x)) = \frac{1}{\sqrt{1-x^{2}}}$$

$$\frac{d}{dx}(\tan^{-1}(x))​ = \frac{1}{1+x^{2}}$$

$$\frac{d}{dx}(sec^{-1}) = \frac{1}{|x|\sqrt{x^{2}-1}$$

We will reverse these to compute integrals.

Important Inverse Trig Integral Formulas

1. Arcsine Integral

$$\int\frac{1}{\sqrt{1-u^{2}}}dx = \sin^{-1}(u)+C$$​

2. Arctangent Integral

$$\int\frac{1}{1+u^{2}}dx = \tan^{-1}(u)+C$$​​

3. Arcsecant Integral

$$\int\frac{1}{u\sqrt{u^{2}-1}dx = \sec^{-1}(u)+C$$​​​

These three forms are the foundation for all inverse trig integrals.

Example 1: Arcsine Type

$$\int\frac{1}{\sqrt{9-x^{2}}}$$

Rewrite:

$$\int\frac{1}{\sqrt{9-x^{2}}} = \int\frac{1}{3\sqrt{1-\frac{x^{2}}{9}}}dx =\frac{1}{3}\int\frac{1}{\sqrt{1-\left(\frac{x}{3}\right)^{2}}}$$

next,use u-substitution

$$​ u=\frac{x}{3} \longrightarrow \frac{1}{3}dx = du \longrightarrow dx = 3 du $$

then

$$\frac{1}{3}\int\frac{1}{\sqrt{1-\left(\frac{x}{3}\right)^{2}}} = \int\frac{1}{\sqrt{1-u^{2}}}du = \sin^{-1}(u)+C = \sin^{-1}(\frac{x}{3})+C$$

Example 2: Arctangent Type

$$\int\frac{1}{4+x^{2}}dx$$

Rewrite:

$$\int\frac{1}{4+x^{2}}dx = \int\frac{1}{4\left(1+\frac{1}{\frac{x^{2}}{4}}\right}dx=\frac{1}{4}\int\frac{1}{1+\left(\frac{x}{2}\right)^{2}}dx$$

next,use u-substitution

$$u=\frac{x}{2} \longrightarrow \frac{du}{dx}=\frac{1}{2} \longrightarrow 2du=dx$$

then

$$\frac{1}{4}\int\frac{1}{1+\left(\frac{x}{2}\right)^{2}}dx=\frac{1}{2}\int\frac{1}{1+u^{2}}du =\frac{1}{2}\tan^{-1}(u)+C = \frac{1}{2}\tan^{-1}(\frac{x}{2})+C$$

Example 3: Arcsecant Type

$$\int\frac{1}{x\sqrt{x^{2}-16}}$$

Rewrite:

$$\int\frac{1}{x\sqrt{x^{2}-16}} = \int\frac{1}{4x\sqrt{\frac{x^{2}}{16}-1}}dx = \int\frac{1}{4x\sqrt{\frac{\left(x}{4}\right)^{2}-1}}dx$$​

next,use u-substitution

$$u=\frac{x}{4} \longrightarrow \frac{du}{dx}=\frac{1}{4} \longrightarrow \frac{1}{4}du=dx and 4u=x$$

then

$$ \int\frac{1}{4x\sqrt{\frac{\left(x}{4}\right)^{2}-1}}dx = \frac{1}{4}\int\frac{1}{16u\sqrt{u^{2}-1}}du = \frac{1}{64}\int\frac{1}{u\sqrt{u^{2}}-1}du = \frac{1}{64}\sec^{-1}+C = \frac{1}{64}\sec^{-1}\left(\frac{x}{4}\right)+C$$

we can find out inverse trigonometry integrals use integration by parts and u-substitution

1 $$\int\sin^{-1}(x)dx$$

$$=\int1\cdot\sin^{-1}(x)dx$$ and set

$$u=\sin^{-1}(x) \longrightarrow du=\frac{1}{\sqrt{1-x^{2}}} , dv=1 \longrightarrow v = x$$ then rewrite,

$$\int\sin^{-1}(x)dx = x\cdot\sin^{-1}(x)-\int\frac{x}{\sqrt{1-x^{2}}}$$ now, u-substitution

$$1-x^{2}=u \longrightarrow -2x dx = du \longrightarrow x dx = -\frac{1}{2}du$$ then

$$x\cdot\sin^{-1}(x)+\frac{1}{2}\int u^{-\frac{1}{2}}du=x\cdot\sin^{-1}(x)+\frac{1}{2}\cdot2u^{\frac{1}{2}}+C=x\cdot\sin^{-1}(x)+\sqrt{1-x^{2}}+C$$

2 $$\int\cos^{-1}(x)dx$$

$$=\int1\cdot\cos^{-1}(x)dx$$ and set

$$u=\cos^{-1}(x) \longrightarrow du=\frac{-1}{\sqrt{1-x^{2}}} , dv=1 \longrightarrow v = x$$ then rewrite,

$$\int\cos^{-1}(x)dx = x\cdot\cos^{-1}(x)+\int\frac{x}{\sqrt{1-x^{2}}}$$ now, u-substitution

$$1-x^{2}=u \longrightarrow -2x dx = du \longrightarrow x dx = -\frac{1}{2}du$$ then

$$x\cdot\cos^{-1}(x)-\frac{1}{2}\int u^{-\frac{1}{2}}du=x\cdot\cos^{-1}(x)-\frac{1}{2}\cdot2u^{\frac{1}{2}}+C=x\cdot\cos^{-1}(x)-\sqrt{1-x^{2}}+C$$

3 $$\int\tan^{-1}(x)dx$$

$$=\int1\cdot\tan^{-1}(x)dx$$ and set

$$u=\tan^{-1}(x) \longrightarrow du=\frac{1}{1+x^{2}} , dv=1 \longrightarrow v = x$$ then rewrite,

$$\int\tan^{-1}(x)dx = x\cdot\tan^{-1}(x)-\int\frac{x}{1+x^{2}}}$$ now, u-substitution

$$1+x^{2}=u \longrightarrow 2x dx = du \longrightarrow x dx = \frac{1}{2}du$$ then

$$x\cdot\tan^{-1}(x)-\frac{1}{2}\int \frac{1}{u}du=x\cdot\tan^{-1}(x)-\frac{1}{2}\cdot\ln(u)+C=x\cdot\tan^{-1}(x)-\frac{1}{2}\ln(1+x^{2})+C$$

4 $$\int\cot^{-1}(x)dx$$

$$=\int1\cdot\cot^{-1}(x)dx$$ and set

$$u=\cot^{-1}(x) \longrightarrow du=\frac{-1}{1+x^{2}} , dv=1 \longrightarrow v = x$$ then rewrite,

$$\int\cot^{-1}(x)dx = x\cdot\cot^{-1}(x)+\int\frac{x}{1+x^{2}}}$$ now, u-substitution

$$1+x^{2}=u \longrightarrow 2x dx = du \longrightarrow x dx = \frac{1}{2}du$$ then

$$x\cdot\cot^{-1}(x)+\frac{1}{2}\int \frac{1}{u}du=x\cdot\cot^{-1}(x)+\frac{1}{2}\cdot\ln(u)+C=x\cdot\cot^{-1}(x)+\frac{1}{2}\ln(1+x^{2})+C$$

5 $$\int\sec^{-1}(x)dx$$

$$=\int1\cdot\sec^{-1}(x)dx$$ and set

$$u=\sec^{-1}(x) \longrightarrow du=\frac{1}{\sqrt{x(x^{2}-1)}} , dv=1 \longrightarrow v = x$$ then rewrite,

$$\int\sec^{-1}(x)dx = x\cdot\sec^{-1}(x)-\int\frac{x}{1+x^{2}}}$$ now, u-substitution

$$x=\sec\theta \longrightarrow dx=\sec\theta\tan\theta and \sqrt{x^{2}+1}=\tan\theta$$ then

$$x\cdot\sec^{-1}(x)-\int \frac{\sec\theta\tan\theta}{\tan\theta}d\theta=x\cdot\sec^{-1}(x)-\int\sec\theta d\theta=x\cdot\sec^{-1}(x)-\ln|\sec\theta+\tan\theta|+C$$

$$=x\cdot\sec^{-1}(x)-ln|x+\sqrt{x^{2}+1}|+C$$

6 $$\int\csc^{-1}(x)dx$$

$$=\int1\cdot\csc^{-1}(x)dx$$ and set

$$u=\csc^{-1}(x) \longrightarrow du=\frac{-1}{\sqrt{x(x^{2}-1)}} , dv=1 \longrightarrow v = x$$ then rewrite,

$$\int\csc^{-1}(x)dx = x\cdot\csc^{-1}(x)+\int\frac{x}{1+x^{2}}}$$ now, u-substitution

$$x=\sec\theta \longrightarrow dx=\sec\theta\tan\theta and \sqrt{x^{2}+1}=\tan\theta$$ then

$$x\cdot\csc^{-1}(x)+\int \frac{\sec\theta\tan\theta}{\tan\theta}d\theta=x\cdot\csc^{-1}(x)+\int\sec\theta d\theta=x\cdot\sec^{-1}(x)+\ln|\sec\theta+\tan\theta|+C$$

$$=x\cdot\csc^{-1}(x)+ln|x+\sqrt{x^{2}+1}|+C$$