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 $

Frequently asked Questions

(1)What are inverse trigonometric functions?
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.

(2)Why are inverse trigonometric functions important in integration?
Answer They are important because they appear in the solutions of integrals involving trigonometric expressions, especially in cases where direct integration is not straightforward. They help in solving integrals related to geometric and physical problems.

(3)Are there any special formulas for integrating inverse trigonometric functions?
Answer Yes, there are several standard formulas, such as:

• $\int \frac{dx}{\sqrt{a^2 – x^2}} = \arcsin(\frac{x}{a}) + C$
• $\int \frac{dx}{a^2 + x^2} = \frac{1}{a}\arctan(\frac{x}{a}) + C$