The value of $\int\frac{dx}{\sin^6x}$ is:
Answer & explanation
Correct answer: option 1
$I=\int\frac{dx}{\sin^6x}=\int cosec^4x.cosec^2x\, dx$ [Put cot x - t ⇒ -cosec2x dx = dt]
$=\int(\cot^2x+1)^2cosec^2x\,dx=\int(t^2+1)^2(-dt)=-\frac{t^5}{5}-2\frac{t^3}{3}-t+C$ (where t = cot x)