How can I evaluate the following integral? ∫cos⁡{x}1+sin⁡{2x}dx I tried the following way, but I was not able to proceed further: I=∫cos⁡{x}(sin⁡{x}+cos⁡{x})2dx =∫sec⁡{x}(1+tan⁡{x})2dx

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How can I evaluate the following integral?  ∫cos⁡{x}1+sin⁡{2x}dx  I tried the following way, but I was not able to proceed further:  I=∫cos⁡{x}(sin⁡{x}+cos⁡{x})2dx   =∫sec⁡{x}(1+tan⁡{x})2dx

Jude Carpenter · Accepted Answer

Hint:    Write numerator as   2cos⁡x=cos⁡x+sin⁡x+(cos⁡x−sin⁡x)  I=∫cos⁡x+sin⁡x{(cos⁡x+sin⁡x})2dx+∫cos⁡x−sin⁡x(cos⁡x+sin⁡x)2dx   I=I1+I2   I1=∫1cos⁡x+sin⁡xdx   I1=∫12(cos⁡x⋅cos⁡{π4+sin⁡x⋅sin⁡{π4})}dx   I1=∫sec⁡(x−π4)2dx   I1=12⋅ln⁡[(sec⁡(x−π4)+tan⁡(x−π4)]+C1   Now   I2=∫cos⁡x−sin⁡x(cos⁡x+sin⁡x)2dx   sin⁡x+cos⁡x=t⇒(cos⁡x−sin⁡x)dx=dt   I2=∫1t2dt   I2=−1t+C2=−1sin⁡x+cos⁡x+C2

How can I evaluate the following integral? \int\frac{\cos{x}}{1+\sin{2x}} dx I tried the

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