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JEE Main · 2024 · 01 Feb 2024 · Shift 2 · Definite Integrals

If

\int_{0}^{\frac{π}{3}} cos^{4} x d x = aπ + b 3

, where

a

and

b

are rational numbers, then

9 a + 8 b

is equal to :

(A)2✓ Correct
(B)1
(C)3
(D) $\frac{3}{2}$

Answer: (A)

Solution

cos^{4} x = (\frac{1 + cos 2 x}{2})^{2} = \frac{1 + 2 cos 2 x + cos ^{2} 2 x}{4} .

Using

cos^{2} 2 x = \frac{1 + cos 4 x}{2}

:

cos^{4} x = \frac{1}{4} + \frac{1}{2} cos 2 x + \frac{1}{8} + \frac{1}{8} cos 4 x = \frac{3}{8} + \frac{1}{2} cos 2 x + \frac{1}{8} cos 4 x .

An antiderivative is

\frac{3 x}{8} + \frac{sin 2 x}{4} + \frac{sin 4 x}{32} .

Evaluating from

0

to

\frac{π}{3}

:

x-term:

\frac{3}{8} \cdot \frac{π}{3} = \frac{π}{8}

.

sin 2 x

term:

\frac{sin ( 2 π /3 )}{4} = \frac{3 /2}{4} = \frac{3}{8}

.

sin 4 x

term:

\frac{sin ( 4 π /3 )}{32} = \frac{- 3 /2}{32} = - \frac{3}{64}

.

Total:

\frac{π}{8} + \frac{3}{8} - \frac{3}{64} = \frac{π}{8} + \frac{7 3}{64}

.

So

a = \frac{1}{8}, b = \frac{7}{64}

, and

9 a + 8 b = \frac{9}{8} + \frac{7}{8} = \frac{16}{8} = 2.

JEE Main · 2024 · 01 Feb 2024 · Shift 2 · Single correct option

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