Question

Find the integrals of the function: $\frac {1}{sin \ x\ cos^3 x}$

Answer 1

$\frac {1}{sin \ x\ cos^3 x}$

= $\frac {sin^2x+cos^2x}{sin \ x\ cos^3 x}$

= $\frac {sin\ x}{cos^3 x}$ + $\frac {1}{sin \ x .\cos \ x}$

= tan x sec2x + $\frac {1/cos^2 x}{sin x .cos x/cos^2 x}$

= tan x sec2x + $\frac {sec^2 x}{tan \ x}$

∴ $\int\frac {1}{sin \ x\ cos^3 x}dx$ = $∫tan \ x .sec^2 x \ dx$+ $∫\frac {sec^2 x}{tan \ x} dx$

Let tan x = t ⇒ sec2x dx = dt
⇒ $\int\frac {1}{sin \ x\ cos^3 x}dx$ = $∫tdt$ + $∫\frac 1t dt$

= $\frac {t^2}{2 }$+ $log|t| +C$

=$\frac 12 tan^2 x +log|tan x|+C$