What's the integral of int (tanx)^5*(secx)^4dx?

1 Answer
Mar 16, 2018

I=tan^6x/6+tan^8x/8+c

Explanation:

We know that,
color(red)(int[f(x)]^n*f^'(x)dx=[f(x)]^(n+1)/(n+1)+c)
f(x)=tanx=>f^'(x)=sec^2x
So,
I=int(tanx)^5*(secx)^4dx=int(tanx)^5(sec^2x)(secx)^2dx
I=int(tanx)^5(1+tan^2x)(sec^2x)dx
I=int(tanx)^5sec^2xdx+int(tanx)^7sec^2xdx
I=int(tanx)^5d/(dx)(tanx)dx+int(tanx)^7d/(dx)(tanx)dx
I=(tanx)^(5+1)/(5+1)+(tanx)^(7+1)/(7+1)+c
I=tan^6x/6+tan^8x/8+c