(a) Evaluate: $\int_{\pi / 6}^{\pi / 3} \frac{\sin x+\cos x}{\sqrt{\sin 2 x}}\, dx$ (b) Show that $\int_{0}^{\pi / 2}(\sqrt{\tan x}+\sqrt{\cot x})\, dx=\sqrt{2}\, \pi$

(a) $2 \sin^{-1}\frac{\sqrt{3}-1}{2}$; (b) Proved

Definite Integrals — CBSE Class 12 · 2008 Board Question with Solution

(a) Evaluate:

\int_{π /6}^{π /3} \frac{s i n x + c o s x}{s i n 2 x} d x

(b) Show that

\int_{0}^{π /2} (tan x + cot x) d x = 2 π

Answer: (a) $2 sin^{- 1} \frac{3 - 1}{2}$ ; (b) Proved

(a) Since

sin 2 x = 1 - (sin x - cos x)^{2}

, put

t = sin x - cos x \Rightarrow d t = (cos x + sin x) d x

.
Limits:

x = \frac{π}{6} \Rightarrow t = \frac{1 - 3}{2}

;

x = \frac{π}{3} \Rightarrow t = \frac{3 - 1}{2}

I = \int_{(1 - 3) /2}^{(3 - 1) /2} \frac{d t}{1 - t ^{2}} = [sin^{- 1} t]_{(1 - 3) /2}^{(3 - 1) /2} = 2 sin^{- 1} \frac{3 - 1}{2}

(the two limits are equal in magnitude and opposite in sign, so the arcsines add).
(b)

tan x + cot x = \frac{sin x + cos x}{sin x cos x} = 2 \cdot \frac{sin x + cos x}{1 - ( sin x - cos x ) ^{2}}

.
Put

t = sin x - cos x

;

x = 0 \to t = - 1

x = \frac{π}{2} \to t = 1

\int_{0}^{π /2} (tan x + cot x) d x = 2 \int_{- 1}^{1} \frac{d t}{1 - t ^{2}} = 2 [sin^{- 1} t]_{- 1}^{1} = 2 \cdot π

Hence shown.

CBSE Class 12 Board · 2008 · Subjective