Evaluate the following integrals: ^5x dx - Vidyadip

Question

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

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Answer

$\int\frac{\cos^5\text{x}}{\sin\text{x}}\text{ dx}$ $=\int\frac{\cos^4\text{x}\cos\text{x}}{\sin\text{x}}\text{ dx}$ $=\int\frac{(\cos^2\text{x})^2\cos\text{x}}{\sin\text{x}}\text{ dx}$ $=\int\frac{(1-\sin^2\text{x})^2\times\cos\text{x}}{\sin\text{x}}\text{ dx}$ $=\int\frac{(1-\sin^4\text{x}-2\sin^2\text{x})}{\sin\text{x}}\cos\text{x}\text{ dx}$Let $\sin\text{x}=\text{t}$
$\Rightarrow\cos\text{x}\text{ dx}=\text{dt}$
Now, $\int\frac{(1-\sin^4\text{x}-2\sin^2\text{x})}{\sin\text{x}}\cos\text{x}\text{ dx}$
$=\int\frac{(1+\text{t}^4-2\text{t}^2)}{\text{t}}\text{ dt}$
$=\int\Big(\frac{1}{\text{t}}+\text{t}^3-2\text{t}\Big)\text{dt}$
$=\log|\text{t}|+\frac{\text{t}^4}{4}-\frac{2\text{t}^2}{2}+\text{C}$
$=\log|\text{t}|+\frac{\text{t}^4}{4}-\text{t}^2+\text{C}$
$=\log|\sin\text{x}|+\frac{\sin^4\text{x}}{4}-\sin^2\text{x}+\text{C}$

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