Evaluate: $\int_{0}^{\pi / 4} \frac{\sin x+\cos x}{16+9 \sin 2 x}\, dx$

CBSE Class 12 · 2018 · Definite IntegralsLong Answer

Evaluate:

\int_{0}^{π /4} \frac{s i n x + c o s x}{16 + 9 s i n 2 x} d x

Answer: $\frac{1}{15} lo g 2$

Put

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

, and

sin 2 x = 1 - t^{2}

, so

16 + 9 sin 2 x = 25 - 9 t^{2}

.
Limits:

x = 0 \to t = - 1

x = \frac{π}{4} \to t = 0

I = \int_{- 1}^{0} \frac{d t}{25 - 9 t ^{2}} = \frac{1}{2 \cdot 5 \cdot 3} [lo g \frac{5 + 3 t}{5 - 3 t}]_{- 1}^{0} = \frac{1}{30} [lo g \frac{5 + 3 t}{5 - 3 t}]_{- 1}^{0}

= \frac{1}{30} [lo g 1 - lo g \frac{2}{8}] = - \frac{1}{30} lo g \frac{1}{4} = \frac{1}{30} lo g 4 = \frac{1}{15} lo g 2

CBSE Class 12 Board · 2018 · Subjective