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

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Answer 1 · 2015-10-15T14:51:28

$1/9tan^9x+C$

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

$1/9tan^9x+C$

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