Question
Evaluate the following intregals:
$\int\frac{1}{\sin^2\text{x}-\sin2\text{x}}\ \text{dx}$

Answer

Let $\text{I}=\int\frac{1}{\sin^2\text{x}-\sin(2\text{x})}\ \text{dx}$
$=\int\frac{1}{\sin^2\text{x}+2\sin\text{x}\cos\text{x}}\text{ dx}$
Dividing numerator and denominator by $\cos^2\text{x}$
$\text{I}=\int\frac{\sec^2\text{x}}{\tan\text{x}+2\tan\text{x}}\ \text{dx}$
Let $\tan\text{x}=\text{t}$
$\sec^2\text{x dx}=\text{dt}$
$\therefore\text{I}=\int\frac{\text{dt}}{\text{t}^2+2\text{t}}$
$=\int\frac{\text{dt}}{\text{t}^2+2\text{t}+1-1}$
$=\ln\frac{\text{dt}}{(\text{t}+1)^2-(-1)^2}$
$=\frac{1}{2}\ln\Big|\frac{\text{t}}{\text{t}+2}\Big|+\text{C}$
$=\frac{1}{2}\ln\Big|\frac{\tan\text{x}}{\tan\text{x}+2}\Big|+\text{C}$

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