Step by step solution to "Evaluate the following integrals. ∫0π2sin7xdx"

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Answered question

2021-10-08

Evaluate the following integrals.

\int_{0}^{\frac{π}{2}} \sin^{7} x d x

Ezra Herbert

Skilled2021-10-09Added 99 answers

Given

\int_{0}^{\frac{π}{2}} \sin^{7} x d x

\sin^{7} (x) = \sin^{6} (x) \sin (x)

and

\sin^{6} (x) = {(\sin^{2} (x))}^{3}

and

\sin^{2} (x) = 1 - \cos^{2} (x)

\Rightarrow \int_{0}^{\frac{π}{2}} \sin^{6} (x) \sin (x) d x

= \int_{0}^{\frac{π}{2}} {(\sin^{2} (x))}^{3} \sin x d x

= \int_{0}^{\frac{π}{2}} {(1 - \cos^{2} (x))}^{3} \sin x d x

Apply u-substitution

u = \cos (x) \Rightarrow d u = - \sin x d x

\Rightarrow - d u = \sin x d x

when

x = 0, u = \cos (0) = 1

when

x = \frac{π}{2}, u = \cos (\frac{π}{2}) = 0

Apply the formula

{(a - b)}^{3} = a^{3} - b^{3} - 3 a^{2} b + 3 a b^{2}

= \int_{0}^{1} [{({(1)}^{3} - (u^{2}))}^{3} - 3 {(1)}^{2} (u^{2}) + 3 (1) {(u^{2})}^{2}] d u

= \int_{0}^{1} (1 - u^{6} - 3 u^{2} + 3 u^{4}) d u

= (1 - 0) - \frac{1}{7} (1 - 0) + \frac{3}{5} (1 - 0)

= 1 - \frac{1}{7} - 1 + \frac{3}{5} = \frac{- 1}{7} + \frac{3}{5}

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