Evaluate the following integrals: ^7x dx - Vidyadip

Question

Evaluate the following integrals:
$\int\sin^7\text{x}\text{ dx}$

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Answer

Let $\text{I}=\int\sin^7\text{x}\text{ dx}$ Then
$\text{I}=\int\sin^6\text{x }\sin\text{x}\text{ dx}$
$=\int\big(\sin^2\text{x}\big)^3\sin\text{x}\text{ dx}$
$=\int\big(1-\cos^2\text{x}\big)^3\sin\text{x}\text{ dx}$
$=\int\big(1-\cos^6\text{x}+3\cos^4\text{x}-3\cos^2\text{x}\big)\sin\text{x}\text{ dx}$
$\text{I}=\int\sin\text{x}\text{ dx}-\int\cos^6\text{x }\sin\text{x}\text{ dx}+3\int\cos^4\text{x }\sin\text{x}\text{ dx}-3\int\cos^2\text{x }\sin\text{x}\text{ dx}$
Putting $\cos\text{x}=\text{t}$ and $-\sin\text{x}\text{dx}=\text{dt}$ in 2^nd, 3^rd and 4^th integral we get
$\text{I}=\int\sin\text{x}\text{ dx}-\int\text{t}^6(-\text{dt})+3\int\text{t}^4(-\text{dt})-3\int\text{t}^2(-\text{dt})$
$=-\cos\text{x}+\frac{\text{t}^7}{7}-\frac{3}{5}\text{t}^5+\frac{3}{3}\text{t}^3+\text{C}$
$=-\cos\text{x}+\frac{\cos^7\text{x}}{7}-\frac{3}{5}\cos^5\text{x}+\cos^3\text{x}+\text{C}$
$\therefore\ \text{I}=-\cos\text{x}+\cos^3\text{x}-\frac{3}{5}\cos^5\text{x}+\frac{1}{7}\cos^7\text{x}+\text{C}$

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