Question

Question #581c1

Answer 1

`int (sin^6(x) + cos^6(x))/ (sin^2(x) cos ^2(x))\ dx=tan(x)-cot(x)-3x+C`

Explanation:

First note that `sin^6(x)+cos^6(x)` is a sum of two cubes `(sin^2(x))^3+(cos^2(x))^3` so it can be factored using `a^3+b^3=(a+b)(a^2-ab+b^2)` and the Pythagorean identity to get

`sin^6(x)+cos^6(x)=(sin^2(x))^3+(cos^2(x))^3`

`=(sin^2(x)+cos^2(x))(sin^4(x)-sin^2(x)cos^2(x)+cos^4(x))`

`=sin^4(x)-sin^2(x)cos^2(x)+cos^4(x)`.

Therefore,

`(sin^6 x + cos^6 x)/ (sin^2(x) cos ^2(x))= (sin^4(x)-sin^2(x)cos^2(x)+cos^4(x))/ (sin^2(x) cos ^2(x))`

`=tan^2(x)-1+cot^2(x)`

But `tan^2(x)=sec^2(x)-1` and `cot^2(x)=csc^2(x)-1`. Hence,

`(sin^6 x + cos^6 x)/ (sin^2(x) cos ^2(x))=sec^2(x)+csc^2(x)-3` and thus

`int (sin^6(x) + cos^6(x))/ (sin^2(x) cos ^2(x))\ dx=int sec^2(x)+csc^2(x)-3\ dx`

`=tan(x)-cot(x)-3x+C`