Question

Evaluate : \[ \int_{0}^{1} \frac{x \tan^{-1} x}{(1 + x^2)^{3/2}} dx \]

Hint:

• Substitution \( x = \tan \theta \) is standard for expressions involving \( \sqrt{1+x^2} \).
• After substitution, simplify trigonometric expressions as much as possible before integrating by parts.

Answer 1

Concept:

• Integration by substitution: \( x = \tan \theta \).
• Integration by parts: \( \int u v dx = u \int v dx - \int (u' \int v dx) dx \).

Step 1: Apply trigonometric substitution
Let \( x = \tan \theta \implies dx = \sec^2 \theta d\theta \)
When \( x = 0, \theta = 0 \). When \( x = 1, \theta = \frac{\pi}{4} \).
The integral becomes:
\( I = \int_{0}^{\pi/4} \frac{\tan \theta \cdot \theta}{(\sec^2 \theta)^{3/2}} \sec^2 \theta d\theta \)
\( I = \int_{0}^{\pi/4} \frac{\tan \theta \cdot \theta}{\sec^3 \theta} \sec^2 \theta d\theta = \int_{0}^{\pi/4} \theta \cdot \frac{\tan \theta}{\sec \theta} d\theta \)
\( I = \int_{0}^{\pi/4} \theta \sin \theta d\theta \)

Step 2: Integrate by parts
Using \( u = \theta, dv = \sin \theta d\theta \):
\( I = [ \theta (-\cos \theta) ]_0^{\pi/4} - \int_{0}^{\pi/4} 1 \cdot (-\cos \theta) d\theta \)
\( I = [ -\theta \cos \theta + \sin \theta ]_0^{\pi/4} \)

Step 3: Evaluate the definite integral
\( I = \left( -\frac{\pi}{4} \cos\frac{\pi}{4} + \sin\frac{\pi}{4} \right) - (0 + 0) \)
\( I = -\frac{\pi}{4} \cdot \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \frac{1}{\sqrt{2}} \left( 1 - \frac{\pi}{4} \right) \)
\( I = \frac{4 - \pi}{4\sqrt{2}} \)