integral of (1 + cos 4x) / cot x + sin x

I'll b thankfull if u show me how 2 solve it!!

I = int ( (1 + cos4x) / (cotx + sinx) ) . dx

I = 2 int ( (cos²2x) ( - sinx) / (cos²x - cosx  - 1) ) . dx

Put y = cosx

so, dy = - sinx . dx

Now, I = 2 int ( (2y² -1)² / ( y² - y - 1 ) . dy

I = 2 int ( (4y²)(y² - -y - 1 + y) + 1) / ( y² - y -1 ) ) . dy

I = 8 int y² . dy + 8 int ( y³ / ( y² - y - 1 ) ) . dy + 2 int ( 1 / ( y² - y - 1 ) ) . dy

I = 8 int y² . dy + 8 int ( 2y + 1 / ( y² - y - 1 ) ) . dy + 8 int (y + 1) . dy+ 2 int ( 1 / ( y² - y - 1 ) ) . dy

I = 8 int y² . dy + 8 int ( 2y - 1 / ( y² - y - 1 ) ) . dy + 2 int ( 1 / ( y² - y - 1 ) ) . dy + 8 int (y + 1) . dy+ 2 int ( 1 / ( y² - y - 1 ) ) . dy

Now, you can integrate. If you still have any problem nudge me.