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

Question

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

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Answer

$\frac{1}{\sin\text{x}\cos^3\text{x}}=\frac{\sin^2\text{x}+\cos^2\text{x}}{\sin\text{x}\cos^3\text{x}}$
$=\frac{\sin\text{x}}{\cos^3\text{x}}+\frac{1}{\sin\text{x}\cos\text{x}}$
$=\tan\text{x}\sec^2\text{x}+\frac{\frac{1\cos^2\text{x}}{\sin\text{x}\cos\text{x}}}{\cos^2\text{x}}$
$=\tan\text{x}\sec^2\text{x}+\frac{\sec^2\text{x}}{\tan\text{x}}$
$\therefore\ \int\frac{1}{\sin\text{x}\cos^3\text{x}}\text{ dx}=\int\tan\text{x}\sec^2\text{x}\text{ dx}+\int\frac{\sec^2\text{x}}{\tan\text{x}}\text{ dx}$
Let $\tan\text{x}=\text{t}$
$\sec^2\text{x}\text{ dx}=\text{dt}$
$\int\frac{1}{\sin\text{x}\cos^3\text{x}}\text{ dx}=\int\text{t}\text{ dt}+\int\frac{1}{\text{t}}\text{ dt}$
$=\frac{\text{t}^2}{2}+\log|\text{t}|+\text{C}$
$=\frac{1}{2}\tan^2\text{x}+\log|\tan\text{x}|+\text{C}$

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