Answer to Question #103214 in Calculus for Patrick Kariuki

Question

Integrate tan³x dx

Answer 1

$\int \tan^3x\ dx =$

$\int \cfrac{\sin^2x\ \sin x}{\cos^3x}\ dx =$

$\int \cfrac{1 - \cos^2x}{\cos^3x} \sin x\ dx =$

$\int \cfrac{\cos^2x - 1}{\cos^3x}\ d(\cos x) =$

$\int \cfrac{d(\cos x)}{\cos x} - \int \cfrac{d(\cos x)}{\cos^3x} =$

$\ln|\cos x| + \cfrac1{2\cos^2x} + C$