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

amevaa0y

amevaa0y

Answered question

2022-01-27

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

Answer & Explanation

Jude Carpenter

Jude Carpenter

Beginner2022-01-28Added 9 answers

Hint:
Write numerator as
2cosx=cosx+sinx+(cosxsinx)
I=cosx+sinx{(cosx+sinx})2dx+cosxsinx(cosx+sinx)2dx
I=I1+I2
I1=1cosx+sinxdx
I1=12(cosxcos{π4+sinxsin{π4})}dx
I1=sec(xπ4)2dx
I1=12ln[(sec(xπ4)+tan(xπ4)]+C1
Now
I2=cosxsinx(cosx+sinx)2dx
sinx+cosx=t(cosxsinx)dx=dt
I2=1t2dt
I2=1t+C2=1sinx+cosx+C2

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