Hi,

 

Look,

 

I am not using the integration sign. Assume it is present there.

 

sin2x / ( sin⁶x + cos⁶x )

As, a³ + b³ f = (a+b)(a² +b² - ab ) 

sin2x / [ (sin²x + cos²x )(sin⁴x + cos⁴x - sin²xcos²x ) ]

As, sin²x + cos²x = 1

sin2x / [ (sin²x + cos²x )² - 2sin²xcos²x - sin²xcos²x ] 

sin2x / [ 1 - 3sin²xcos²x ]

Divide num and denom by cos⁴x. And yes, open sin2x.

2Tanx*sec²x / [ sec⁴x - 3Tan²x ]

Write (Tan²x + 1)²  in place of  sec⁴x. Open it.

 

Now, put t = tanx [ substitution and find the answer, ans !!! ].

 

Hope you have got it !!!