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int (sin2x)/(sin^4x+cos^4x) dx...

`int (sin2x)/(sin^4x+cos^4x) dx`

Text Solution

Verified by Experts

Let `I=∫(sin2x)/(sin^4x+cos^4x)​dx`
Dividing the numerator and denominator by `cos^4x`
`I=∫(2tanx.sec^2x)/(tan^4x+1)​dx`
Let `tanx=u`
`(sec^2x) dx=du`
So,
`I=∫(2u​du)/(1+u^4)`
Let `u^2=z`
...
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