Q.Integrate the function $\sin^{-1}\left(\frac{2x}{1+x^2}\right)$

The key idea is to simplify the inverse trigonometric expression using a standard substitution before integrating.

Let $x = \tan \theta$, so that $\theta = \tan^{-1} x$. Then

Hence $\sin^{-1}\left(\frac{2x}{1+x^2}\right) = \sin^{-1}(\sin 2\theta)$. For the principal range, this equals $2\theta = 2\tan^{-1} x$ when $|x| \le 1$.

The integral becomes

$\int 2\tan^{-1} x \, dx.$ …

We start with the integral:

Why This Approach Works

The expression $\frac{2x}{1+x^2}$ is a dead giveaway for the tangent double-angle identity. Recall that $\tan(2\theta) = \frac{2\tan\theta}{1-\tan^2\theta}$, but here we have a plus sign in the denominator. That points to the sine double-angle identity in a different form: if $x = \tan \theta$, then $\sin(2\theta) = \frac{2\tan\theta}{1+\tan^2\theta} = \frac{2x}{1+x^2}$. So the integrand becomes $\sin^{-1}(\sin(2\theta))$, which simplifies to $2\theta$ — but only for a specific range of $\theta$ where the inverse sine is well-defined. This substitution turns a messy inverse trigonometric function into a simple algebraic one.

Step-by-Step Solution

1. Substitute $x = \tan \theta$. Then $dx = \sec^2 \theta \, d\theta$, and the integrand becomes:

(for $\theta \in [-\pi/4, \pi/4]$). So the integral transforms to:

2. Apply integration by parts. Let $u = \theta$ and $dv = \sec^2 \theta \, d\theta$. Then $du = d\theta$ and $v = \tan \theta$. Using $\int u \, dv = uv - \int v \, du$:

3. Integrate $\tan \theta$. …

Method: Simplify an inverse-trig integrand with a trig substitution first

Some inverse-trig integrands collapse to a simple multiple of $\tan^{-1}x$ (or $\sin^{-1}x$) once you substitute $x=\tan\theta$ and use a standard double-angle identity — simplify before you integrate.

Steps

Step 1: Recognise the tell-tale form.

$\dfrac{2x}{1+x^{2}}$, $\dfrac{2x}{1-x^{2}}$, $\dfrac{1-x^{2}}{1+x^{2}}$ are the $\sin$, $\tan$, $\cos$ of $2\theta$ when $x=\tan\theta$.

Step 2: Substitute $x=\tan\theta$ and simplify the inverse-trig.

Here $\dfrac{2x}{1+x^{2}}=\sin 2\theta$, so $\sin^{-1}\!\left(\dfrac{2x}{1+x^{2}}\right)=2\theta=2\tan^{-1}x$ (for $|x|\le 1$).

Step 3: Integrate the simplified function. …

Common Mistakes

Mistake 1: Integrating $\sin^{-1}\!\left(\dfrac{2x}{1+x^{2}}\right)$ head-on.

Why it's wrong: it is needlessly hard; the argument is $\sin 2\theta$ with $x=\tan\theta$. Correct approach: simplify to $2\tan^{-1}x$ first.

Mistake 2: Writing $2\tan^{-1}x$ as valid for all $x$. …