₀²π} ^5≤ft( x/4 ) {d}x is equal to

Correct Answer is : 3215{32}{{15}}1532

Correct Answer is : 3215{32}{{15}}1532

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UPSC NDA II 2015 Mathematics Paper

Question : 64 of 120

Marks: +1, -0

\int_{0}^{2\pi} \sin^5\left( \frac{x}{4} \right) \mathrm{d}x

is equal to

Solution:

Concept:

Use the identity

\sin^2\theta + \cos^2\theta = 1

and substitution to evaluate the definite integral.

Explanation:

Let

I = \int_{0}^{2\pi} \sin^5\left(\frac{x}{4}\right) \, dx

Rewrite as

\int_{0}^{2\pi} \sin^2\left(\frac{x}{4}\right) \sin^2\left(\frac{x}{4}\right) \sin\left(\frac{x}{4}\right) \, dx

Replace

\sin^2\left(\frac{x}{4}\right)

with

1 - \cos^2\left(\frac{x}{4}\right)

Then

I = \int_{0}^{2\pi} \left(1 - \cos^2\left(\frac{x}{4}\right)\right)^2 \sin\left(\frac{x}{4}\right) \, dx

Substitute

t = \cos\left(\frac{x}{4}\right)

so that

dt = -\frac{1}{4}\sin\left(\frac{x}{4}\right) dx

\Rightarrow

\sin\left(\frac{x}{4}\right) dx = -4\, dt

When

x = 0

t = \cos 0 = 1

; when

x = 2\pi

t = \cos\left(\frac{\pi}{2}\right)=0

Thus

I = \int_{1}^{0} (1 - t^2)^2 (-4) \, dt = 4 \int_{0}^{1} (1 - t^2)^2 \, dt

(reversing limits removes the negative).

Expand

(1 - t^2)^2 = 1 - 2t^2 + t^4

Integrate:

4 \left[ t - \frac{2t^3}{3} + \frac{t^5}{5} \right]_{0}^{1} = 4 \left( 1 - \frac{2}{3} + \frac{1}{5} \right) = 4 \left( \frac{15 - 10 + 3}{15} \right) = 4 \cdot \frac{8}{15} = \frac{32}{15}

Answer:

\frac{32}{15}

(Option C).

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