Question

How do you find the integral of `int tan^8 x *sec^2 x dx`?

Answer 1

`1/9tan^9x+C`

Explanation:

Using the substitution technique, we may let `u=tanx`
Then `du=sec^2xdx`

Now rewriting the original integral in terms of u, we get :

`intu^8du=u^9/9+C`

Now converting back to in terms of x, we get the final answer as

`1/9tan^9x+C`