Answer to Question #175018 in Calculus for Phyroe

Question

∫ csc⁴ θ dθ

Answer 1

Solution:

$\int \csc^4 \theta\ d\theta \\=\int \csc^2 \theta \csc^2 \theta\ d\theta \\=\int \csc^2 \theta (1+\cot^2 \theta)\ d\theta \\=\int \csc^2 \theta \ d\theta +\int \csc^2 \theta\cot^2 \theta\ d\theta$

$\\=-\cot \theta-\dfrac{(\cot \theta)^3}{3}+C$ [$\because \int [f(x)]^nf'(x)dx=\dfrac{[f(x)]^{n+1}}{n+1}+C$ ]

$\\=-\cot \theta-\dfrac{\cot^3 \theta}{3}+C$