Integration of Inverse Trigonometric Functions

The integration of inverse trigonometric functions involves finding the antiderivatives of the six inverse trigonometric functions: 

• sin^-1 x (arcsin x)
• cos^-1 x (arccos x)
• tan^-1 x (arctan x)
• cot^-1 x (arccot x)
• sec^-1 x (arcsec x) 
• cosec^-1 x (arccosec x) 

Formulas

Here are the standard integration formulas for the six inverse trigonometric functions:

Inverse Trigonometric Functions	Integrals
${\int \sin ^{-1}xdx}$	${x\sin ^{-1}x+\sqrt{1-x^{2}}+C}$
${\int \cos ^{-1}xdx}$	${x\cos ^{-1}x-\sqrt{1-x^{2}}+C}$
${\int \tan ^{-1}xdx}$	${x\tan ^{-1}x-\dfrac{1}{2}\ln \left| 1+x^{2}\right| +C}$
${\int \cot ^{-1}xdx}$	${x\cot ^{-1}x+\dfrac{1}{2}\ln \left| 1+x^{2}\right| +C}$
${\int \sec ^{-1}xdx}$	${x\sec ^{-1}x-\ln \left| x+\sqrt{x^{2}-1}\right| +C}$
${\int \text{cosec}\, ^{-1}xdx}$	${x\text{cosec}\, ^{-1}x+\ln \left| x+\sqrt{x^{2}-1}\right| +C}$

Proofs 

For Inverse Sine Function 

We need to verify: 

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

Solving: 

Here, we 

Here, ${\int \sin ^{-1}xdx}$

Step 1. Using Integration By-Parts

= ${\int \sin ^{-1}x\cdot 1dx}$

By using the by-parts formula of integration, we get

= ${\sin ^{-1}x\int dx-\int \left[ \dfrac{d}{dx}\left( \sin ^{-1}x\right) \int 1dx\right] dx}$

Since ${\dfrac{d}{dx}\left( \sin ^{-1}x\right) =\dfrac{1}{\sqrt{1-x^{2}}}}$

= ${\sin ^{-1}x\left( x\right) -\int \left[ \dfrac{1}{\sqrt{1-x^{2}}}\left( x\right) \right] dx}$

= ${x\sin ^{-1}x-\int \dfrac{x}{\sqrt{1-x^{2}}}dx}$ …..(i)

Step 2. Substituting 

Now, let us consider 1 – x² = u 

Differentiating both sides with respect to x, we get

⇒ -2x dx = du

⇒ x dx = ${-\dfrac{1}{2}du}$

Substituting the values in the expression (i), we get

= ${x\sin ^{-1}x-\left( -\dfrac{1}{2}\right) \int \dfrac{1}{\sqrt{u}}du}$

= ${x\sin ^{-1}x+\dfrac{1}{2}\int \dfrac{1}{\sqrt{u}}du}$

By using the power rule of the integration, we get

= ${x\sin ^{-1}x+\dfrac{1}{2}\left( 2\sqrt{u}\right) +C}$

= ${x\sin ^{-1}x+\sqrt{u}+C}$

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

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

For Inverse Cosine Function

We need to verify:

${\int \cos ^{-1}xdx}$ = ${x\cos ^{-1}x-\sqrt{1-x^{2}}+C}$

Solving:

Here, ${\int \cos ^{-1}xdx}$

Step 1. Using Integration By-Parts

= ${\int \cos ^{-1}x\cdot 1dx}$

By using the by-parts formula of integration, we get

= ${\cos ^{-1}x\int dx-\int \left[ \dfrac{d}{dx}\left( \cos ^{-1}x\right) \int 1dx\right] dx}$

Since ${\dfrac{d}{dx}\left( \cos ^{-1}x\right) =-\dfrac{1}{\sqrt{1-x^{2}}}}$

= ${\cos ^{-1}x\left( x\right) -\int \left[ -\dfrac{1}{\sqrt{1-x^{2}}}\left( x\right) \right] dx}$

= ${x\cos ^{-1}x+\int \dfrac{x}{\sqrt{1-x^{2}}}dx}$ …..(i)

Step 2. Substituting 

Now, let us consider 1 – x² = u 

Differentiating both sides with respect to x, we get

⇒ -2x dx = du

⇒ x dx = ${-\dfrac{1}{2}du}$

Substituting the values in the expression (i), we get

= ${x\cos ^{-1}x+\left( -\dfrac{1}{2}\right) \int \dfrac{1}{\sqrt{u}}du}$

= ${x\cos ^{-1}x-\dfrac{1}{2}\int \dfrac{1}{\sqrt{u}}du}$

By using the power rule of the integration, we get

= ${x\cos ^{-1}x-\dfrac{1}{2}\left( 2\sqrt{u}\right) +C}$

= ${x\cos ^{-1}x-\sqrt{u}+C}$

= ${x\cos ^{-1}x-\sqrt{1-x^{2}}+C}$

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

For Inverse Tangent Function

We need to verify:

${\int \tan ^{-1}xdx}$ = ${x\tan ^{-1}x-\dfrac{1}{2}\ln \left| 1+x^{2}\right| +C}$

Solving:

Here, ${\int \tan ^{-1}xdx}$

Step 1. Using Integration By-Parts

= ${\int \tan ^{-1}x\cdot 1dx}$

By using the by-parts formula of integration, we get

= ${\tan ^{-1}x\int dx-\int \left[ \dfrac{d}{dx}\left( \tan ^{-1}x\right) \int 1dx\right] dx}$

Since ${\dfrac{d}{dx}\left( \tan ^{-1}x\right) =\dfrac{1}{1+x^{2}}}$

= ${\tan ^{-1}x\left( x\right) -\int \left[ \dfrac{1}{1+x^{2}}\left( x\right) \right] dx}$

= ${x\tan ^{-1}x-\int \dfrac{x}{1+x^{2}}dx}$ …..(i)

Step 2. Substituting 

Now, let us consider 1 + x² = u 

Differentiating both sides with respect to x, we get

⇒ 2x dx = du

⇒ x dx = ${\dfrac{1}{2}du}$

Substituting the values in the expression (i), we get

= ${x\tan ^{-1}x-\left( \dfrac{1}{2}\right) \int \dfrac{1}{u}du}$

By using the power rule of the integration, we get

= ${x\tan ^{-1}x-\dfrac{1}{2}\ln \left| u\right| +C}$

= ${x\tan ^{-1}x-\dfrac{1}{2}\ln \left| 1+x^{2}\right| +C}$

Thus, ${\int \tan ^{-1}xdx}$ = ${x\tan ^{-1}x-\dfrac{1}{2}\ln \left| 1+x^{2}\right| +C}$

For Inverse Cotangent Function

We need to verify:

${\int \cot ^{-1}xdx}$ = ${x\cot ^{-1}x+\dfrac{1}{2}\ln \left| 1+x^{2}\right| +C}$

Solving:

Here, ${\int \cot ^{-1}xdx}$

Step 1. Using Integration By-Parts

= ${\int \cot ^{-1}x\cdot 1dx}$

By using the by-parts formula of integration, we get

= ${\cot ^{-1}x\int dx-\int \left[ \dfrac{d}{dx}\left( \cot ^{-1}x\right) \int 1dx\right] dx}$

Since ${\dfrac{d}{dx}\left( \cot ^{-1}x\right) =-\dfrac{1}{1+x^{2}}}$

= ${\cot ^{-1}x\left( x\right) -\int \left[ -\dfrac{1}{1+x^{2}}\left( x\right) \right] dx}$

= ${x\cot ^{-1}x+\int \dfrac{x}{1+x^{2}}dx}$ …..(i)

Step 2. Substituting

Now, let us consider 1 + x² = u 

Differentiating both sides with respect to x, we get

⇒ 2x dx = du

⇒ x dx = ${\dfrac{1}{2}du}$

Substituting the values in the expression (i), we get

= ${x\cot ^{-1}x+\left( \dfrac{1}{2}\right) \int \dfrac{1}{u}du}$

By using the power rule of the integration, we get

= ${x\cot ^{-1}x+\dfrac{1}{2}\ln \left| u\right| +C}$

= ${x\cot ^{-1}x+\dfrac{1}{2}\ln \left| 1+x^{2}\right| +C}$

Thus, ${\int \cot ^{-1}xdx}$ = ${x\cot ^{-1}x+\dfrac{1}{2}\ln \left| 1+x^{2}\right| +C}$

For Inverse Secant Function

We need to verify:

${\int \sec ^{-1}xdx}$ = ${x\sec ^{-1}x-\ln \left| x+\sqrt{x^{2}-1}\right| +C}$

Solving:

Here, ${\int \sec ^{-1}xdx}$

Step 1. Using Integration By-Parts

= ${\int \sec ^{-1}x\cdot 1dx}$

By using the by-parts formula of integration, we get

= ${\sec ^{-1}x\int dx-\int \left[ \dfrac{d}{dx}\left( \sec ^{-1}x\right) \int 1dx\right] dx}$

Since ${\dfrac{d}{dx}\left( \sec ^{-1}x\right) =\dfrac{1}{\left| x\right| \sqrt{x^{2}-1}}}$

= ${\sec ^{-1}x\left( x\right) -\int \left[ \dfrac{1}{x}\sqrt{x^{2}-1}\left( x\right) \right] dx}$

= ${x\sec ^{-1}x-\int \dfrac{1}{\sqrt{x^{2}-1}}dx}$ …..(i)

Step 2. Substituting

Now, let us consider x = sec u 

Differentiating both sides with respect to x, we get

⇒ dx = sec u tan u du

Also, ${\sqrt{x^{2}-1}}$ = ${\sqrt{\sec ^{2}u-1}}$ = tan u

Substituting the values in the expression (i), we get

= ${x\sec ^{-1}x-\int \dfrac{\sec u\tan u}{\tan u}du}$

= ${x\sec ^{-1}x-\int \sec udu}$

Since ${\int \sec xdx=\ln \left| \sec x+\tan x\right| +C}$

= ${x\sec ^{-1}x-\ln \left| \sec u+\tan u\right| +C}$

= ${x\sec ^{-1}x-\ln \left| \sec u+\sqrt{\sec ^{2}u-1}\right| +C}$

= ${x\sec ^{-1}x-\ln \left| x+\sqrt{x^{2}-1}\right| +C}$

Thus, ${\int \sec ^{-1}xdx}$ = ${x\sec ^{-1}x-\ln \left| x+\sqrt{x^{2}-1}\right| +C}$

For Inverse Cosecant Function

We need to verify:

${\int \text{cosec}\, ^{-1}xdx}$${x\text{cosec}\, ^{-1}x+\ln \left| x+\sqrt{x^{2}-1}\right| +C}$

Solving:

Here, ${\int \text{cosec}\, ^{-1}xdx}$

Step 1. Using Integration By-Parts

= ${\int \text{cosec}\, ^{-1}x\cdot 1dx}$

By using the by-parts formula of integration, we get

= ${\text{cosec}\, ^{-1}x\int dx-\int \left[ \dfrac{d}{dx}\left( \text{cosec}\, ^{-1}x\right) \int 1dx\right] dx}$

Since ${\dfrac{d}{dx}\left( \text{cosec}\, ^{-1}x\right) =-\dfrac{1}{\left| x\right| \sqrt{x^{2}-1}}}$

= ${\text{cosec}\, ^{-1}x\left( x\right) -\int \left[ -\dfrac{1}{x}\sqrt{x^{2}-1}\left( x\right) \right] dx}$

= ${x\text{cosec}\, ^{-1}x+\int \dfrac{1}{\sqrt{x^{2}-1}}dx}$ …..(i)

Step 2. Substituting

Now, let us consider x = sec u 

Differentiating both sides with respect to x, we get

⇒ dx = sec u tan u du

Also, ${\sqrt{x^{2}-1}}$ = ${\sqrt{\sec ^{2}u-1}}$ = tan u

Substituting the values in the expression (i), we get

= ${x\text{cosec}\, ^{-1}x+\int \dfrac{\sec u\tan u}{\tan u}du}$

= ${x\text{cosec}\, ^{-1}x+\int \sec udu}$

Since ${\int \sec xdx=\ln \left| \sec x+\tan x\right| +C}$

= ${x\text{cosec}\, ^{-1}x+\ln \left| \sec u+\tan u\right| +C}$

= ${x\text{cosec}\, ^{-1}x+\ln \left| \sec u+\sqrt{\sec ^{2}u-1}\right| +C}$

= ${x\text{cosec}\, ^{-1}x+\ln \left| x+\sqrt{x^{2}-1}\right| +C}$

Thus, ${\int \text{cosec}\, ^{-1}xdx}$ = ${x\text{cosec}\, ^{-1}x+\ln \left| x+\sqrt{x^{2}-1}\right| +C}$

Integrals Resulting in Inverse Trigonometric Functions

Some integrals naturally result in inverse trigonometric functions. These include: ${\int \dfrac{1}{\sqrt{a^{2}-x^{2}}}dx}$, ${\int -\dfrac{1}{\sqrt{a^{2}-x^{2}}}dx}$, and ${\int \dfrac{1}{a^{2}+x^{2}}dx}$, which correspond to arc sin, arccos, and arctan respectively.

Here is a list of the expressions involving the 6 trigonometric functions and their corresponding integrals:

Integrals	Inverse Trigonometric Functions
${\int \dfrac{1}{\sqrt{a^{2}-x^{2}}}dx}$	${\sin ^{-1}\dfrac{x}{a}+C}$
${\int -\dfrac{1}{\sqrt{a^{2}-x^{2}}}dx}$	${\cos ^{-1}\dfrac{x}{a}+C}$
${\int \dfrac{1}{a^{2}+x^{2}}dx}$	${\dfrac{1}{a}\tan ^{-1}\dfrac{x}{a}+C}$
${\int -\dfrac{1}{a^{2}+x^{2}}dx}$	${\dfrac{1}{a}\cot ^{-1}\dfrac{x}{a}+C}$
${\int \dfrac{1}{x\sqrt{x^{2}-a^{2}}}dx}$	${\dfrac{1}{a}\sec ^{-1}\dfrac{x}{a}+C}$
${\int -\dfrac{1}{x\sqrt{x^{2}-a^{2}}}dx}$	${\dfrac{1}{a}\text{cosec}\, ^{-1}\dfrac{x}{a}+C}$

Formulas for these functions are directly derived from derivatives of inverse trigonometric functions. For example,

Proof of ${\int \dfrac{1}{\sqrt{a^{2}-x^{2}}}dx}$

Let us consider y = ${\sin ^{-1}\dfrac{x}{a}}$

Step 1. Substituting 

⇒ a sin y = x

Step 2. Differentiating Both Sides

By differentiating with respect to x, we get

⇒ ${\dfrac{d}{dx}\left( a\sin y\right) =\dfrac{d}{dx}\left( x\right)}$

⇒ ${a\cos y\dfrac{dy}{dx}=1}$

⇒ ${\dfrac{dy}{dx}=\dfrac{1}{a\cos y}}$ …..(i)

For ${-\dfrac{\pi }{2}\leq y\leq \dfrac{\pi }{2}}$, cos y ≥ 0

Step 3. Expressing in Terms of x

Now, by applying the Pythagorean identity sin² y + cos² y = 1

⇒ cos y = ${\sqrt{1-\sin ^{2}y}}$

Now, from (i), we get

⇒ ${\dfrac{dy}{dx}=\dfrac{1}{a\cos y}}$

⇒ ${\dfrac{dy}{dx}=\dfrac{1}{a\sqrt{1-\sin ^{2}y}}}$

⇒ ${\dfrac{dy}{dx}=\dfrac{1}{\sqrt{a^{2}-a^{2}\sin ^{2}y}}}$

⇒ ${\dfrac{dy}{dx}=\dfrac{1}{\sqrt{a^{2}-x^{2}}}}$ …..(ii)

Step 4. Integrating Both Sides

Then, for -a ≤ x ≤ a, integrating both sides of the equation (ii), we get

⇒ ${\int dy=\int \dfrac{1}{\sqrt{a^{2}-x^{2}}}dx}$

⇒ y = ${\int \dfrac{1}{\sqrt{a^{2}-x^{2}}}dx}$

Since y = ${\sin ^{-1}\dfrac{x}{a}}$

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

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

Similarly, we can also verify the formulas of the other integrals. 

Solved Examples