Evaluate the following integrals: 1 ^4x ^2x dx - Vidyadip

Question

Evaluate the following integrals:
$\int\frac{1}{\sin^4\text{x}\cos^2\text{x}}\text{dx}$

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Answer

$\int\frac{1}{\sin^4\text{x}\cos^2\text{x}}\text{dx}$
Dividing numerator & denominator by $\sin^2\text{x}$
$=\int\frac{\frac{1}{\sin^2\text{x}}}{\sin^4\text{x }\cdot\ \cot^2\text{x}}\text{dx}$
$=\int\frac{\text{cosec}^6\text{x}}{\cot^2}\text{dx}$
$=\int\frac{\text{cosec}^4\text{x }\cdot\text{ cosec}^2\text{x}\text{ dx}}{\cot^2\text{x}}$
$=\int\frac{(1+\cot^2\text{x})^2\text{cosec}^2\text{x}\text{ dx}}{\cot^2\text{x}}$
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, $\int\frac{(1+\cot^2\text{x})^2\text{cosec}^2\text{x}\text{ dx}}{\cot^2\text{x}}$
$=\int\Big(\frac{1+\text{t}^2}{\text{t}}\Big)^2(-\text{dt})$
$=-\int\frac{\big(1+\text{t}^4+2\text{t}^2\big)}{\text{t}^2}\text{dt}$
$=-\int(\text{t}^{-2}+\text{t}^2+2)\text{dt}$
$=-\Big[\frac{\text{t}^{-2+1}}{-2+1}+\frac{\text{t}^3}{3}+2\text{t}\Big]+\text{C}$
$=-\Big[-\frac{1}{\text{t}}+\frac{\text{t}^3}{3}+2\text{t}\Big]+\text{C}$
$=-\frac{1}{3}\text{t}^3-2\text{t}+\frac{1}{\text{t}}+\text{C}$
$=-\frac{1}{3}\cot^3\text{x}-2\cot\text{x}+\frac{1}{\cot\text{x}}+\text{C}$
$=-\frac{1}{3}\cot^3\text{x}-2\cot\text{x}+\tan\text{x}+\text{C}$

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