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# Integration of inverse trigonometric functions

Integration of inverse trigonometric functions can be calculated using integration by parts ,integration by substitution .Here are the formula for it

$\int \sin^{-1}x \, dx = x \sin^{-1}x + \sqrt{1 – x^2} + C$

$\int \cos^{-1}x \, dx = x \cos^{-1}x – \sqrt{1 – x^2} + C$

$\int \tan^{-1}x \, dx = x \tan^{-1}x – \frac {1}{2} \ln (1 + x^2) + C$

$\int \csc^{-1}x \, dx = x \csc^{-1}x + \ln |x + \sqrt { x^2+ 1}| + C$

$\int \sec^{-1}x \, dx = x \sec^{-1}x – \ln |x + \sqrt { x^2- 1}| + C$

$\int \cot^{-1}x \, dx = x \cot^{-1}x + \frac {1}{2} \ln (1 + x^2) + C$

## Proof of sin inverse

let $\theta = \sin^{-1}x$
$sin \theta = x$
$cos \theta d \theta = dx$

$$\int \sin^{-1}x \, dx = \int \theta cos \theta d \theta$$

Now solving this using integration parts

$= \theta \int cos \theta d \theta – \int (\frac {d}{d \theta } \theta ) .(\int cos \theta d \theta) \; \theta \\ = \theta \sin \theta – \int \sin \theta \; d \theta \\ =\theta \sin \theta + \cos \theta$

Substituting back the values

$\int \sin^{-1}x \, dx = x \sin^{-1}x + \sqrt {1-x^2} + C$

## Proof of cos inverse

let $\theta = \cos^{-1}x$
$cos \theta = x$
$-sin \theta d \theta = dx$

$$\int \cos^{-1}x \, dx = -\int \theta sin \theta d \theta$$

Now solving this using integration parts

$= -\theta \int sin \theta d \theta + \int (\frac {d}{d \theta } \theta ) .(\int sin \theta d \theta) \; \theta \\ = \theta \cos \theta – \int \cos \theta \; d \theta \\ =\theta \cos \theta – \sin \theta$

Substituting back the values

$\int \cos^{-1}x \, dx = x \cos^{-1}x – \sqrt {1-x^2} + C$

## Proof of tan inverse

To find the integral we use integration by parts. Integration by parts is based on the product rule for differentiation and is given by:

$\int f(x) g(x) dx = f(x) (\int g(x) dx )- \int \left \{ \frac {df(x)}{dx} \int g(x) dx \right \} dx$
In our case, we can let $f(x) = \tan^{-1}x$ and $g(x) = 1$. Then

• $\frac {df(x)}{dx} = \frac{1}{1+x^2} \, dx$
• $\int g(x) dx = \int dx = x$

Now, substitute these into the integration by parts formula:

$$\int \tan^{-1}x \, dx = x \tan^{-1}x – \int x \cdot \frac{1}{1+x^2} \, dx$$

$=x \tan^{-1}x – \int \frac{x}{1+x^2} \, dx \\ =x \tan^{-1}x – \frac {1}{2} \int \frac{2x}{1+x^2} \, dx$

Now lets calculate the second integral separately $\int \frac{2x}{1+x^2} \, dx$

Let $t= 1+x^2$
then $dt=2x dx$
Therefore

$\int \frac{2x}{1+x^2} \, dx = \int \frac{1}{t} \, dx \\ = \ln |t| = \ln |1+x^2|$

Substituting this value in main integral , we get

$$\int \tan^{-1}x \, dx = x \tan^{-1}x – \frac {1}{2} \ln (1 + x^2) + C$$

## Proof of cosec inverse

let $\theta = \csc^{-1}x$
$\csc \theta = x$
$-\csc \theta cot \theta d \theta = dx$

$$\int \csc^{-1}x \, dx = \int \theta (-\csc \theta cot \theta) d \theta$$

Now solving this using integration parts

$= \theta \int (-\csc \theta cot \theta) d \theta – \int (\frac {d}{d \theta } \theta ) .(\int (-\csc \theta cot \theta)) \; \theta \\ = \theta \csc \theta – \int \csc \theta \; d \theta$

Now

$\int \csc(x) \, dx = \ln | \csc(x) – \cot(x) | + C$

Therefore

$=\theta \csc \theta – \ln | \csc(x) – \cot(x) | + C$

Now $cot \theta= \sqrt {\csc^2 \theta -1} = \sqrt {x^2-1}$

Substituting back the values

$\int \csc^{-1}x \, dx = x \csc^{-1}x – \ln |x – \sqrt {x^2 -1}|+ C$
$= x \csc^{-1}x + \ln |\frac {1}{x – \sqrt {x^2 -1}}| + C$
$= x \csc^{-1}x + \ln |x + \sqrt {x^2 -1}| + C$

## Proof of sec inverse

let $\theta = \sec^{-1}x$
$\sec \theta = x$
$\sec \theta tan \theta d \theta = dx$

$$\int \sec^{-1}x \, dx = \int \theta (\sec \theta tan \theta) d \theta$$

Now solving this using integration parts

$= \theta \int (\sec \theta tan \theta) d \theta – \int (\frac {d}{d \theta } \theta ) .(\int (\sec \theta tan \theta)) \; \theta \\ = \theta \sec \theta – \int \sec \theta \; d \theta$

Now

$\int \sec(x) \, dx = \ln | \sec(x) + \tan(x) | + C$

Therefore

$=\theta \sec \theta – \ln | \sec(x) + \tan(x) | + C$

Now $tan \theta= \sqrt {\sec^2 \theta -1} = \sqrt {x^2-1}$

Substituting back the values

$\int \sec^{-1}x \, dx = x \sec^{-1}x – \ln |x + \sqrt {x^2 -1}|+ C$

## Proof of cot inverse

To find the integral we use integration by parts. Integration by parts is based on the product rule for differentiation and is given by:

$\int f(x) g(x) dx = f(x) (\int g(x) dx )- \int \left \{ \frac {df(x)}{dx} \int g(x) dx \right \} dx$
In our case, we can let $f(x) = \cot^{-1}x$ and $g(x) = 1$. Then

• $\frac {df(x)}{dx} = -\frac{1}{1+x^2} \, dx$
• $\int g(x) dx = \int dx = x$

Now, substitute these into the integration by parts formula:

$$\int \cot^{-1}x \, dx = x \cot^{-1}x – \int x \cdot \frac{-1}{1+x^2} \, dx$$

$=x \tan^{-1}x + \int \frac{x}{1+x^2} \, dx \\ =x \tan^{-1}x – \frac {1}{2} \int \frac{2x}{1+x^2} \, dx$

Now lets calculate the second integral separately $\int \frac{2x}{1+x^2} \, dx$

Let $t= 1+x^2$
then $dt=2x dx$
Therefore

$\int \frac{2x}{1+x^2} \, dx = \int \frac{1}{t} \, dx \\ = \ln |t| = \ln |1+x^2|$

Substituting this value in main integral , we get

$$\int \cot^{-1}x \, dx = x \cot^{-1}x + \frac {1}{2} \ln (1 + x^2) + C$$

## Integrals that result in Inverse Trigonometric Functions

Integrals that result in inverse trigonometric functions, like $\int \frac{dx}{\sqrt{1 – x^2}} = \sin^{-1}(x) + C$ or $\int \frac{dx}{1 + x^2} = \tan^{-1}(x) + C$.

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

$\int (\frac {1}{\sqrt {1-x^2}}) = – \cos ^{-1}x +C$

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

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

$\int (\frac {1}{|x|\sqrt {x^-1}}) = -sec^{-1} x + C$

$\int (\frac {1}{|x|\sqrt {x^-1}}) = -cosec^{-1} x + C$

Answer Inverse trigonometric functions are functions that reverse the action of the trigonometric functions. They include $sin^{-1}(x)$, $cos^{-1}(x)$, $tan^{-1}(x)$, $cot^{-1}(x)$, $sec^{-1}(x)$, and $csc^{-1}(x)$, and are used to find the angle whose trigonometric function equals a given number.