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Definite Integration

Section: Mathematics

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Question : 1 of 54

Marks: +1, -0

\int\limits_{-2\pi}^{2\pi} \sin^4(2x) \cos^6(2x) dx=

[AP EAPCET 24 May 2025 S1]

$\frac{3\pi}{64}$
$\frac{9\pi}{64}$
$\frac{9\pi}{35}$
$\frac{9\pi}{280}$

Solution:

I=\int\limits_{-2\pi}^{2\pi} \sin^4(2x) \cos^6(2x) dx

Let

u=2x

, then

du=2dx

,

so

dx=\frac{1}{2} du

When

x=-2\pi, u=2(-2\pi)=-4\pi

when

x=2\pi, u=2(2\pi)=4\pi

I=\int\limits_{-2\pi}^{2\pi} \sin^4(2x) \cos^6(2x) dx

=\frac{1}{2} \int\limits_{-4\pi}^{4\pi} \sin^4(u) \cos^6(u) du

Since,

\sin(u)

and

\cos(u)

are periodic with a period of

2\pi

,

So,

\sin^4(u)

and

\cos^6(u)

are also periodic with a period of

2\pi

=\frac{1}{2} \cdot 2 \int\limits_{-2\pi}^{2\pi} \sin^4(u) \cos^6(u) du

=\int\limits_{-2\pi}^{2\pi} \sin^4(u) \cos^6(u) du

=\frac{1}{2} \cdot 4 \int\limits_{0}^{2\pi} \sin^4(u) \cos^6(u) du

=2 \int\limits_{0}^{2\pi} \sin^4(u) \cos^6(u) du

Since, the function,

f(u)=\sin^4(u) \cos^6(u)

is an even function.

\therefore 2 \int\limits_{0}^{2\pi} \sin^4(u) \cos^6(u) du

=2 \cdot 2 \int\limits_{0}^{\pi} \sin^4(u) \cos^6(u) du

=4 \int\limits_{0}^{\pi} \sin^4(u) \cos^6(u) du

=4 \cdot 2 \int\limits_{0}^{\pi/2} \sin^4(u) \cos^6(u) du

=8 \int\limits_{0}^{\pi/2} \sin^4(u) \cos^6(u) du

We know that,

\int\limits_{0}^{\pi/2} \sin^m(u) \cos^n(u) du

=\frac{(m-1)(m-3)\dots (1\ \text{or}\ 2)(n-1)(n-3)\dots (1\ \text{or}\ 2)}{(m+n)(m+n-2)\dots (1\ \text{or}\ 2)} \times \alpha

where,

\alpha=\frac{\pi}{2}

if both

m

and

n

are even, and

\alpha=1

otherness.

Here,

m=4, n=6

both are even.

So,

\int\limits_{0}^{\pi/2} \sin^4(4) \cos^6(4) du

=\frac{(4-1)(4-3)(6-1)(6-3)(6-5)}{(4+6)(4+6-2)(4+6-4)} \cdot \frac{\pi}{2}

(4+6-6)(4+6-8)

=\frac{3 \cdot 1 \cdot 5 \cdot 3 \cdot 1}{10 \cdot 8 \cdot 6 \cdot 4 \cdot 2} \cdot \frac{\pi}{2} = \frac{45}{3840} \cdot \frac{\pi}{2} = \frac{3}{256} \cdot \frac{\pi}{2}

=\frac{3\pi}{512}

\therefore 8 \int\limits_{0}^{\pi/2} \sin^4(u) \cos^6(u) du

=8 \cdot \frac{3\pi}{512} = \frac{24\pi}{512} = \frac{3\pi}{64}

\therefore 8 \int\limits_{0}^{\frac{\pi}{2}} \sin^4(u) \cos^6(u) du = \frac{3\pi}{64}

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