Q.Integrate the function $\tan^{-1}\sqrt{\frac{1-x}{1+x}}$

Concept: Power Rule Integration — after a trigonometric substitution, the integrand simplifies to a standard form.

Let $x = \cos 2\theta$, so $\sqrt{\frac{1-x}{1+x}} = \sqrt{\frac{1-\cos 2\theta}{1+\cos 2\theta}} = \sqrt{\frac{2\sin^2\theta}{2\cos^2\theta}} = \tan\theta$.

Then $\tan^{-1}(\tan\theta) = \theta$ (for principal values), and $dx = -2\sin 2\theta \, d\theta$.

The integral becomes:

Integrate by parts: let $u = \theta$, $dv = \sin 2\theta \, d\theta$, so $du = d\theta$, $v = -\frac{1}{2}\cos 2\theta$.

Then: …

The key simplification

Let $x=\cos 2\theta$. Then

so $\sqrt{\frac{1-x}{1+x}} = \tan\theta$ and the integrand becomes $\tan^{-1}(\tan\theta)=\theta$ (for $\theta\in[0,\tfrac{\pi}{2})$). Since $\theta=\tfrac{1}{2}\cos^{-1}x$, the integral reduces to $\tfrac{1}{2}\int\cos^{-1}x\,dx$.

Step-by-step solution

1. Reduce the integral.

2. Integrate by parts with $u=\cos^{-1}x$, $dv=dx$:

3. Combine. …

Method: Trig substitution to simplify an inverse-trig integrand, then integrate by parts

Use this when the integrand is an inverse trig function of a surd like $\tan^{-1}\sqrt{\tfrac{1-x}{1+x}}$: a substitution such as $x=\cos 2\theta$ turns the whole thing into a simple angle.

Steps

Step 1: Substitute to unwrap the inverse trig.

Put $x=\cos 2\theta$. Then $\dfrac{1-x}{1+x}=\tan^2\theta$, so $\sqrt{\tfrac{1-x}{1+x}}=\tan\theta$ and $\tan^{-1}(\tan\theta)=\theta=\tfrac12\cos^{-1}x$. …

Common Mistakes

Mistake 1: Trying to integrate $\tan^{-1}\sqrt{\tfrac{1-x}{1+x}}$ directly.

Why it's wrong: the nested surd inside an inverse-trig has no direct antiderivative; a substitution is needed first. Correct approach: put $x=\cos 2\theta$ to reduce it to $\tfrac12\cos^{-1}x$.

Mistake 2: Wrong half-angle reduction.

Why it's wrong: $\dfrac{1-\cos2\theta}{1+\cos2\theta}=\tan^2\theta$; a slip here breaks the simplification. Correct approach: use $1-\cos2\theta=2\sin^2\theta$, $1+\cos2\theta=2\cos^2\theta$. …