Question

Find the integrals of the function: $tan^4 x$

Answer 1

tan⁴x = tan²x × tan²x

= (sec²x-1) × tan²x [ as we know, sec²x - tan²x = 1 ]

= sec²x × tan²x - tan²x

= sec²x tan²x- (sec²x-1)

= sec²x tan²x - sec²x+1

now, $\int{tan^4x}\,dx=\int{sec^2x.tan^2x-sec^2x+1}\,dx$

= $\int{sec^2x.tan^2x}\,dx-\int{sec^2x}\,dx+\int{1}\,dx$

now, Let $I_1=\int{sec^2x.tan^2x}\,dx$

Let tanx = t

differentiating both sides,

sec²x dx = dt

so, $\int{sec^2x.tan^2x}=\int{t^2}\,dt$

= $\left[\frac{t^3}{3}\right]$

hence, $I_1=\left[\frac{tan^3x}{3}\right]$

now, let, $I_2=\int{sec^2x}\,dx=tanx$

so, $\int{tan^4}\,dx=I_1-I_2+x+C$

= $\left[\frac{tan^3}{3}\right]-tanx+x+C$