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JEE Main · 2024 · 31 Jan 2024 · Shift 1 · Definite Integrals

If the integral

525 \int_{0}^{\frac{π}{2}} sin 2 x cos^{\frac{11}{2}} x (1 + cos^{\frac{5}{2}} x)^{\frac{1}{2}} d x

is equal to

(n 2 - 64)

, then

n

is equal to

Answer: 176

Solution

I = \int_{0}^{π /2} sin 2 x cos^{11/2} x (1 + cos^{5/2} x)^{1/2} d x

Write

sin 2 x = 2 sin x cos x

and substitute

t = cos x

(

d t = - sin x d x

):

I = \int_{0}^{1} 2 t^{13/2} (1 + t^{5/2})^{1/2} d t

Let

u = t^{5/2}

, so

d u = \frac{5}{2} t^{3/2} d t

, and

t^{13/2} d t = t^{5} \cdot t^{3/2} d t = u^{2} \cdot \frac{2}{5} d u

:

I = \frac{4}{5} \int_{0}^{1} u^{2} 1 + u d u

Let

w = 1 + u

:

\int_{1}^{2} (w - 1)^{2} w d w = \int_{1}^{2} (w^{5/2} - 2 w^{3/2} + w^{1/2}) d w = [\frac{2}{7} w^{7/2} - \frac{4}{5} w^{5/2} + \frac{2}{3} w^{3/2}]_{1}^{2}

Evaluating (with

2^{7/2} = 82, 2^{5/2} = 42, 2^{3/2} = 22

):

= \frac{44 2 - 16}{105}

So

I = \frac{4}{5} \cdot \frac{44 2 - 16}{105} = \frac{176 2 - 64}{525}

, giving

525 I = 1762 - 64

.

So

n = 176

.

JEE Main · 2024 · 31 Jan 2024 · Shift 1 · Integer answer

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