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TG EAMCET 2 May 2025 Shift 1 Paper

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Question : 78 of 160

Marks: +1, -0

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

$\frac{3\pi}{128}$
$\frac{9\pi}{32}$
$\frac{9\pi}{64}$
$\frac{3\pi}{64}$

Solution:

\int\limits_{-2\pi}^{2\pi} \sin^4 x \cos^6 x \, dx

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

Since,

\sin x

is an odd function and

\sin^2 x

is an even function and

\cos^6 x

is an even function.

So,

f(x)

is an even function.

\therefore \int\limits_{-2\pi}^{2\pi} \sin^4 x \cos^6 x \, dx = 2 \int\limits_{0}^{2\pi} \sin^4 x \cos^6 x \, dx

\left[ \because \int_{-a}^{a} f(x) \, dx = 2 \int\limits_{0}^{a} f(x) \, dx \right]

= 2 \times 2 \int\limits_{0}^{\pi} \sin^4 x \cos^6 x \, dx

\left[ \because \int\limits_{0}^{n\pi} f(x) \, dx = n \int\limits_{0}^{T} f(x) \, dx \right]

= 4 \int\limits_{0}^{\pi} \sin^4 x \cos^6 x \, dx

= 4 \times 2 \int\limits_{0}^{\frac{\pi}{2}} \sin^4 x \cos^6 x \, dx

\left[ \because \int\limits_{0}^{2a} f(x) \, dx = 2 \int\limits_{0}^{a} f(x) \, dx \right]

= 8 \int\limits_{0}^{\frac{\pi}{2}} \sin^4 x \cos^6 x \, dx

\left[ \because \text{ If } m \text{ and } n \text{ are both even, then } \right]

\int\limits_{0}^{\frac{\pi}{2}} \sin^m x \cos^n x \, dx =

\left[ \frac{[(m-1)(m-3) \ldots 2 \text{ or } 1][(n-1)(n-3) \ldots 2 \text{ or } 1]}{[(m+n)(m+n-2) \ldots 2 \text{ or } 1]} \times \frac{\pi}{2} \right]

= 8 \times \frac{3 \times 1 \times 5 \times 3 \times 1}{10 \times 8 \times 6 \times 4 \times 2} \times \frac{\pi}{2}

= \frac{3\pi}{64}

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