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Evaluate: int1/(2-3cos2x)\ dx...

Evaluate: `int1/(2-3cos2x)\ dx`

Text Solution

Verified by Experts

Given,
I=`int1/(2-3cos^2x) dx`
I=`int1/(2-3(1-sin^2x)) dx`
I=`int1/(6sin^2x-1) dx`
dividing numerator denominator by `sec^2x`
`∫(sec^2x)/(5tan^2x−sec^2x) dx​`
`=∫(sec^2x)/(5tan^2x−1) dx​`
` =1/2​∫(dtanx)/(5tan^2x−1)​` ...
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