Evaluate the following integrals: ^3x ^6x dx - Vidyadip

Question

Evaluate the following integrals:
$\int\sin^3\text{x}\cos^6\text{x}\text{ dx}$

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Answer

Let $\text{I}=\int\sin^3\text{x}\cos^6\text{x}\text{ dx}$
Here, power of $\sin\text{x}$ is odd, so we substitute
$\cos\text{x}=\text{t}$
$-\sin\text{x}\text{dt}=\text{dt}$
$\therefore\ \text{I}=\int\sin^2\text{x}\text{ t}^6(-\text{dt})$
$=-\int\big(1-\cos^2\text{x}\big)\text{t}^6\text{dt}$
$=-\int\big(\text{t}^6-\text{t}^8\big)\text{dt}$
$=-\frac{\text{t}^7}{7}+\frac{\text{t}^9}{9}+\text{C}$
$\therefore\ \text{I}=\frac{1}{7}\cos^7\text{x}+\frac{1}{9}\cos^9\text{x}+\text{C}$

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