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JEE Main · 2024 · 04 Apr 2024 · Shift 1 · Definite Integrals

If

\int_{0}^{\frac{π}{4}} \frac{s i n ^{2} x}{1 + s i n x c o s x} d x = \frac{1}{a} lo g_{e} (\frac{a}{3}) + \frac{π}{b 3}

, where

a, b \in N

, then

a + b

is equal to ____.

Answer: 8

Solution

Let

I = \int_{0}^{π /4} \frac{sin ^{2} x}{1 + sin x cos x} d x = \int_{0}^{π /4} \frac{1 - cos 2 x}{2 + sin 2 x} d x

.

Split as

I = I_{1} - I_{2}

with

I_{1} = \int_{0}^{π /4} \frac{d x}{2 + sin 2 x}

and

I_{2} = \int_{0}^{π /4} \frac{cos 2 x}{2 + sin 2 x} d x

.
Using

t = tan x

,

I_{1} = \frac{π}{6 3}

; and

I_{2} = \frac{1}{2} ln \frac{3}{2}

directly.

I = \frac{1}{2} ln \frac{2}{3} + \frac{π}{6 3} (form \frac{1}{a} ln \frac{a}{3} + \frac{π}{b 3}),

so

a = 2, b = 6

and

a + b = 8

.
Answer: 8.

JEE Main · 2024 · 04 Apr 2024 · Shift 1 · Integer answer

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