Using properties of definite integrals, evaluate: $\int_{0}^{\pi} \frac{x \tan x}{\sec x+\tan x} d x$.

CBSE Class 12 · 2008 · Definite IntegralsLong Answer

Using properties of definite integrals, evaluate:

\int_{0}^{π} \frac{x t a n x}{s e c x + t a n x} d x

Answer: $\frac{π}{2} (π - 2)$

Simplify:

\frac{tan x}{sec x + tan x} = \frac{sin x}{1 + sin x}

, so

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

.
Apply

x \to π - x

(with

sin (π - x) = sin x

) and add:

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

.
Write

\frac{sin x}{1 + sin x} = 1 - \frac{1}{1 + sin x}

and rationalise

\frac{1}{1 + sin x} = \frac{1 - sin x}{cos ^{2} x} = sec^{2} x - sec x tan x

2 I = π \int_{0}^{π} [1 - sec^{2} x + sec x tan x] d x = π [x - tan x + sec x]_{0}^{π}

= π [(π - 0 - 1) - (0 - 0 + 1)] = π (π - 2)

Hence

I = \frac{π}{2} (π - 2)

CBSE Class 12 Board · 2008 · Subjective