Evaluate the following integrals: 1 ^3x dx - Vidyadip

Question

Evaluate the following integrals:
$\int\frac{1}{\sin^3\text{x}\cos\text{x}}\text{dx}$

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Answer

$\int\frac{1}{\sin^3\text{x}\cos\text{x}}\text{ dx}$
Dividing numerator & denominator by $\sin^4\text{x}$
$=\int\frac{\frac{1}{\sin^4\text{x}}\text{ dx}}{\frac{\sin^3\text{x}\cdot\cos\text{x}}{\sin^4\text{x}}}$
$=\int\frac{\text{cosec}^4\text{x}\text{ dx}}{\cot\text{x}}$
$=\int\frac{\text{cosec}^2\text{x}\cdot\text{cosec}^2\text{x}\text{ dx}}{\cot\text{x}}$
$=\int\frac{(1+\cot^2\text{x})\cdot\text{cosec}^2\text{x}\text{ dx}}{\cot\text{x}}$
Let $\cot\text{x}=\text{t}$
$\text{cosec}^2\text{x}=\frac{\text{dt}}{\text{dx}}$
$\text{cosec}^2\text{x}\text{dx}=-\text{dt}$
Now, $\int\frac{(1+\cot^2\text{x})\cdot\text{cosec}^2\text{x}}{\cot\text{x}}\text{ dx}$
$=\int\frac{(1+\text{t}^2)\cdot(-\text{dt})}{\text{t}}$
$=-\int\Big(\frac{1}{\text{t}}+\text{t}\Big)\text{dt}$
$=-\log|\text{t}|-\frac{\text{t}^2}{2}+\text{C}$
$=-\log|\cot\text{x}|-\frac{\cot^2\text{x}}{1}+\text{C}$
$=\log|\cot\text{x}|^{-1}-\frac{(\text{cosec}^2\text{x}-1)}{2}+\text{C}$
$=\log\Big|\frac{1}{\cot\text{x}}\Big|-\frac{\text{cosec}^2\text{x}}{2}+\frac{1}{2}+\text{C}$
$=\log|\tan\text{x}|-\frac{1}{2\sin^2\text{x}}+\text{C}'$ $\Big[\therefore\text{C}'=\text{C}+\frac{1}{2}\Big]$

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