Evaluate the following integrals: ^5x dx - Vidyadip

Question

Evaluate the following integrals:
$\int\sin^5\text{x}\text{ dx}$

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Answer

Let $\text{I}=\int\sin^5\text{x}\text{ dx}$ Then
$\text{I}=\int\sin^3\text{x }\sin^2\text{x}\text{ dx}$
$=\int\sin^3\text{x}(1-\cos^2\text{x})\text{ dx} $
$=\int\big(\sin^3\text{x}-\sin^3\text{x}\cos^2\text{x}\big)\text{ dx}$
$=\int\big[\sin\text{x}(1-\cos^2\text{x})-\sin^3\text{x}\cos^2\text{x}\big]\text{dx}$
$=\int\big(\sin\text{x}-\sin\text{x}\cos^2\text{x}-\sin^3\text{x}\cos^2\text{x}\big)\text{dx}$
$\text{I}=\int\sin\text{x}\text{ dx}-\int\sin\text{x}\cos^2\text{x}\text{ dx}-\int\sin^3\text{x}\cos^2\text{x}\text{dx}$
Putting $\cos\text{x}=\text{t}$ and $-\sin\text{x}\text{ dx}=\text{dt}$ is 2^nd and 3^rd integrals, we get
$\text{I}=\int\sin\text{x}\text{ dx}-\int\text{t}^2(-\text{dt})+\int\sin^2\text{x}\text{t}^2\text{ dt}$
$=\int\sin\text{x}\text{ dx}+\int\text{t}^2\text{dt}+\int\big(1-\cos^2\text{x}\big)\text{t}^2\text{dt}$
$=\int\sin\text{x}\text{ dx}+\int\text{t}^2\text{dt}+\int(1-\text{t}^2)\text{t}^2\text{dt}$
$=-\cos\text{x}+\frac{\text{t}^3}{3}+\frac{\text{t}^3}{3}-\frac{\text{t}^5}{5}+\text{C}$
$=-\cos\text{x}+\frac{2}{3}\text{t}^3-\frac{1}{5}\text{t}^5+\text{C}$
$=-\cos\text{x}+\frac{2}{3}(\cos^3\text{x})-\frac{1}{5}\big(\cos^5\text{x})+\text{C}$
$\therefore\ \text{I}=-\big[\cos\text{x}-\frac{2}{3}\cos^3\text{x}+\frac{1}{5}\cos^5\text{x}\big]+\text{C}$

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