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Integrate log(sinx+cosx)dx from -pi/4 to pi/4. someone plz help

Saboor , 9 Years ago
Grade 12
anser 2 Answers
Riddhish Bhalodia

Last Activity: 9 Years ago

This is a tricky question
First put x = -x and add
I = \int \limits_{-\pi/4}^{\pi/4}log(sinx+cosx)dx = \int \limits_{-\pi/4}^{\pi/4}log(-sinx+cosx)dx
2I = \int \limits_{-\pi/4}^{\pi/4}log(-sin^2x+cos^2x)dx = \int \limits_{-\pi/4}^{\pi/4}log(cos2x)dx = 2\int \limits_{0}^{\pi/4}log(cos2x)dxI =\int \limits_{0}^{\pi/4}log(cos2x)dx
Now put 2x = t which implies, dx = dt/2
I =1/2\int \limits_{0}^{\pi/2}log(cost)dt
Now refer to the following steps to solve for the simplified integral

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xfg

Last Activity: 7 Years ago

what is first step?
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