How to solve this integral ??

int((sin^(6)x + cos^(6)x)/(sin^(2)x * cos^(2)x))

1 Answer
Aug 21, 2017

tanx-3x-cotx+C

Explanation:

int(sin^6x+cos^6x)/(sin^2xcos^2x)dx

Firstly, we need to simplify the integral to make it more accessible

So let
" "s=sinx, c=cosx

=int((s^6+c^6)/(s^2c^2))dx

Rewriting and simplifying the integral

(s^6+c^6)/(s^2c^2)=((s^3)^2+(c^3)^2)/(s^2c^2)

The numerator is sum of cubes

Using

a^3+b^3=(a+b)(a^2-ab+b^2)

We have,

((s^2)^3+(c^2)^3)/(s^2c^2)=((s^2+c^2)(s^4-s^2c^2+c^4))/(s^2c^2)

But s^2+c^2=1

:.=((s^4-s^2c^2+c^4))/(s^2c^2)

Now substitute back the sines and cosines

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

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

=sin^2x/cos^2x-1+cos^2x/sin^2x

=tan^2x-1+cot^2x

now back to the integral!

int(sin^6x+cos^6x)/(sin^2xcos^2x)dx

=int(tan^2x-1+cot^2x)dx

=int(sec^2x-1-1+csc^2x-1)dx

=int(sec^2x-3+csc^2x)dx

These are standard integrals!!

=tanx-3x-cotx+C