Question

Evaluate the integral of sec 5θ tan 5θ dθ

Answer 1

Let us evaluate the integral of $\int \sec 5θ \tan 5θ dθ:$

$\int \sec 5θ \tan 5θ dθ=\int \frac{1}{\cos 5θ}\frac{ \sin 5θ}{\cos 5θ} dθ= -\frac{1}{5}\int \frac{1}{\cos^2 5θ} d(\cos 5θ)= -\frac{1}{5}\int \cos^{-2} 5θ d(\cos 5θ)= \frac{1}{5}\ \cos^{-1} 5θ + C=\frac{1}{5\cos 5θ}+C$