Q.Evaluate the integral using substitution $\int_{0}^{1}\sin^{-1}\left(\frac{2x}{1+x^2}\right)dx$

The key idea is to use the substitution $x = \tan \theta$, which simplifies the argument of the inverse sine.

Let $x = \tan \theta$. Then $dx = \sec^2 \theta \, d\theta$. When $x = 0$, $\theta = 0$; when $x = 1$, $\theta = \frac{\pi}{4}$.

The integrand becomes:

since $2\theta \in [0, \frac{\pi}{2}]$ for $\theta \in [0, \frac{\pi}{4}]$, which is within the principal range of $\sin^{-1}$.

The integral transforms to:

Integrate by parts: let $u = 2\theta$, $dv = \sec^2\theta \, d\theta$, so $du = 2\,d\theta$, $v = \tan\theta$. Then:

We are asked to evaluate

The expression inside the inverse sine, $\frac{2x}{1+x^2}$, is a classic double-angle form. If you recall the tangent half-angle identities, you know that for $x = \tan \theta$,

This is the natural path: the substitution $x = \tan \theta$ will simplify the integrand dramatically.

But there is a subtlety: the range of $\sin^{-1}$ is $[-\pi/2, \pi/2]$, and for $x \in [0,1]$, $\theta$ runs from $0$ to $\pi/4$, so $2\theta$ lies in $[0, \pi/2]$, safely inside the principal range. No sign issues.

Let’s work through it step by step.

1. Substitute $x = \tan \theta$. Then $dx = \sec^2 \theta \, d\theta$. When $x = 0$, $\theta = 0$; when $x = 1$, $\theta = \pi/4$. The integral becomes

2. Simplify the argument. Since $1 + \tan^2 \theta = \sec^2 \theta$, we have

Therefore,

3. Handle the inverse sine. For $\theta \in [0, \pi/4]$, $2\theta \in [0, \pi/2]$, and on this interval $\sin^{-1}(\sin 2\theta) = 2\theta$ (since sine is one-to-one and increasing there). So

4. Integrate by parts. Let $u = \theta$ and $dv = \sec^2 \theta \, d\theta$. Then $du = d\theta$ and $v = \tan \theta$. Integration by parts gives

We know $\int \tan \theta \, d\theta = -\log |\cos \theta| + C$, so

5. Evaluate the definite integral.

At $\theta = \pi/4$: $\tan(\pi/4) = 1$, $\cos(\pi/4) = \sqrt{2}/2$, so $\log(\cos(\pi/4)) = \log(1/\sqrt{2}) = -\frac{1}{2} \log 2$.

At $\theta = 0$: $\theta \tan \theta = 0 \cdot 0 = 0$, and $\log(\cos 0) = \log 1 = 0$.

Hence

Method: Trigonometric substitution to simplify an inverse-trig integrand

An argument like $\dfrac{2x}{1+x^2}$ (or $\dfrac{1-x^2}{1+x^2}$) is a disguised double-angle form; substituting $x=\tan\theta$ collapses the inverse-trig function to a plain multiple of $\theta$.

Steps

Step 1: Recognise the double-angle template and substitute $x=\tan\theta$.

Then $dx=\sec^2\theta\,d\theta$, and $\dfrac{2\tan\theta}{1+\tan^2\theta}=\sin 2\theta$.

Step 2: Collapse the inverse function on its valid range.

$\sin^{-1}(\sin 2\theta)=2\theta$ only while $2\theta\in[-\tfrac{\pi}{2},\tfrac{\pi}{2}]$ — always check the limits fall in the principal range before dropping the inverse.

Step 3: Integrate the resulting $\theta\sec^2\theta$ by parts.

Take $u=\theta$, $dv=\sec^2\theta\,d\theta$, giving $\displaystyle\int\theta\sec^2\theta\,d\theta=\theta\tan\theta-\int\tan\theta\,d\theta=\theta\tan\theta+\log|\cos\theta|.$

Step 4: Evaluate at the transformed limits.

Common Mistakes

Mistake 1: Writing $\sin^{-1}(\sin 2\theta)=2\theta$ without checking the range.

Why it's wrong: the identity holds only when $2\theta\in[-\tfrac{\pi}{2},\tfrac{\pi}{2}]$; if the limits pushed $2\theta$ outside this, a correction is needed. Correct approach: confirm $\theta\in[0,\tfrac{\pi}{4}]$ so $2\theta\in[0,\tfrac{\pi}{2}]$ is safe.

Mistake 2: Forgetting the $\sec^2\theta$ from $dx=\sec^2\theta\,d\theta$.

Why it's wrong: the integral is $\int 2\theta\sec^2\theta\,d\theta$, not $\int 2\theta\,d\theta$; dropping $\sec^2\theta$ loses the by-parts entirely. Correct approach: keep $dx=\sec^2\theta\,d\theta$ and integrate $\theta\sec^2\theta$ by parts.