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JEE Main · 2025 · 29 Jan 2025 · Shift 1 · Indefinite Integrals

The integral

80 \int_{0}^{\frac{π}{4}} (\frac{s i n θ + c o s θ}{9 + 16 s i n 2 θ}) d θ

is equal to :

(A) $3 lo g_{e} 4$
(B) $6 lo g_{e} 4$
(C) $4 lo g_{e} 3$ ✓ Correct
(D) $2 lo g_{e} 3$

Answer: (C)

Solution

Write

sin 2 θ = 2 sin θ cos θ

and set

t = sin θ - cos θ

, so

d t = (cos θ + sin θ) d θ

and

t^{2} = 1 - sin 2 θ

, i.e.

sin 2 θ = 1 - t^{2}

. Then

9 + 16 sin 2 θ = 25 - 16 t^{2} .

As

θ

runs from

0

to

π /4

,

t

runs from

- 1

to

0

, so

80 \int_{0}^{π /4} \frac{sin θ + cos θ}{9 + 16 sin 2 θ} d θ = 80 \int_{- 1}^{0} \frac{d t}{25 - 16 t ^{2}} .

Using

\int \frac{d t}{a ^{2} - t ^{2}} = \frac{1}{2 a} ln \frac{a + t}{a - t}

with

a = 5/4

:

80 \cdot \frac{1}{40} [ln (\frac{5/4 + t}{5/4 - t})]_{- 1}^{0} = 2 [ln 1 - ln \frac{1}{9}] = 2 ln 9 = 4 ln 3.

So the integral equals

4 lo g_{e} 3

.

JEE Main · 2025 · 29 Jan 2025 · Shift 1 · Single correct option

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