Question

$\int x\text{Tan}^{-1}\sqrt{\frac{1+x^2}{1-x^2}}dx=$

• A. $\frac{x^2}{4}(\pi-\text{Cos}^{-1}x^2)+\frac{1}{4}\sqrt{1-x^4}+c$
• B. $\frac{x^2}{4}(\pi-\text{Cos}^{-1}x^2)-\frac{1}{4}\sqrt{1-x^4}+c$
• C. $\frac{x^2}{4}(\pi+\text{Cos}^{-1}x^2)-\frac{1}{4}\sqrt{1-x^4}+c$
• D. $\frac{x^2}{4}(\pi+\text{Cos}^{-1}x^2)-\frac{1}{4}\sqrt{1-x^2}+c$

Hint:

For integrals involving terms like $\sqrt{\frac{1 \pm x^2}{1 \mp x^2}}$, the substitution $x^2 = \cos\theta$ is often very effective, as it allows for simplification using half-angle trigonometric identities.

Answer 1

The Correct Option is A

Let's use the substitution $x^2 = \cos\theta$.
Differentiating, $2x dx = -\sin\theta d\theta \implies x dx = -\frac{1}{2}\sin\theta d\theta$.
The integral becomes $\int \text{Tan}^{-1}\sqrt{\frac{1+\cos\theta}{1-\cos\theta}} (x dx)$.
First, simplify the term inside the arctan:
$\sqrt{\frac{1+\cos\theta}{1-\cos\theta}} = \sqrt{\frac{2\cos^2(\theta/2)}{2\sin^2(\theta/2)}} = \sqrt{\cot^2(\theta/2)} = |\cot(\theta/2)|$.
Assuming we are in a domain where $\cot(\theta/2)$ is positive, this is $\cot(\theta/2)$.
$\text{Tan}^{-1}(\cot(\theta/2)) = \text{Tan}^{-1}(\tan(\frac{\pi}{2}-\frac{\theta}{2})) = \frac{\pi}{2}-\frac{\theta}{2}$.
Substitute this back into the integral:
$\int \left(\frac{\pi}{2}-\frac{\theta}{2}\right) \left(-\frac{1}{2}\sin\theta d\theta\right) = -\frac{1}{4}\int (\pi-\theta)\sin\theta d\theta$.
Now we use integration by parts: $\int u dv = uv - \int v du$.
Let $u = \pi-\theta$ and $dv = \sin\theta d\theta$. Then $du = -d\theta$ and $v = -\cos\theta$.
$-\frac{1}{4} \left[ (\pi-\theta)(-\cos\theta) - \int (-\cos\theta)(-d\theta) \right] = -\frac{1}{4} \left[ -(\pi-\theta)\cos\theta - \int \cos\theta d\theta \right]$.
$= -\frac{1}{4} [ -(\pi-\theta)\cos\theta - \sin\theta ] + C = \frac{1}{4}(\pi-\theta)\cos\theta + \frac{1}{4}\sin\theta + C$.
Now substitute back using $x^2 = \cos\theta$, so $\theta = \text{Cos}^{-1}(x^2)$. Also, $\sin\theta = \sqrt{1-\cos^2\theta} = \sqrt{1-x^4}$.
$\frac{1}{4}(\pi-\text{Cos}^{-1}(x^2))(x^2) + \frac{1}{4}\sqrt{1-x^4} + C$.
Rearranging, we get $\frac{x^2}{4}(\pi-\text{Cos}^{-1}x^2)+\frac{1}{4}\sqrt{1-x^4}+c$.