Evaluate: $\int_{0}^{\pi} \frac{x \sin x}{1+\cos ^{2} x} d x$

CBSE Class 12 · 2008 · Definite IntegralsLong Answer

Evaluate:

\int_{0}^{π} \frac{x s i n x}{1 + c o s ^{2} x} d x

Answer: $\frac{π ^{2}}{4}$

Let

I = \int_{0}^{π} \frac{x sin x}{1 + cos ^{2} x} d x

. Apply

x \to π - x

; note

sin (π - x) = sin x

and

cos^{2} (π - x) = cos^{2} x

I = \int_{0}^{π} \frac{( π - x ) sin x}{1 + cos ^{2} x} d x

Adding:

2 I = π \int_{0}^{π} \frac{sin x}{1 + cos ^{2} x} d x

.
Put

t = cos x \Rightarrow d t = - sin x d x

;

x = 0 \to t = 1

x = π \to t = - 1

2 I = π \int_{- 1}^{1} \frac{d t}{1 + t ^{2}} = π [tan^{- 1} t]_{- 1}^{1} = π \cdot \frac{π}{2}

Hence

I = \frac{π ^{2}}{4}

CBSE Class 12 Board · 2008 · Subjective