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## siddarth95 3 years ago Help prove this integral

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1. siddarth95

$\int\limits_{0}^{\pi/2}\frac{ dx }{ \sin x + \cos x } = \frac{ 1 }{ \sqrt{2} } \ln \ \frac{ \sqrt{2}+1 }{ \sqrt{2}-1 }$

2. Jonask

let u=sin x +cos x du=cosx-sinx dx

3. Jonask

$\int \frac{1}{ \cos x+\sin x}dx$ multiply by cos x-sin x both numerator and denominator

4. Jonask

$\int \frac{\cos x-\sin x}{\cos^2x-\sin^2x} dx=\int \frac{\cancel{\cos x-\sin x}du}{\cos 2x}\frac{du}{\cancel{\cos x-\sin x}}$

5. Jonask

$\int \frac{du}{\cos 2x}=\int \sec 2x=\ln|\sec 2x+\tan 2x|$/2

6. Jonask

error i never substituted u

7. Meepi

I don't have time to put an explanation here, but use the substitution u = tan(x/2): http://www-math.mit.edu/~djk/18_01/chapter24/section03.html

8. Meepi

Pretty sure others can help you if you get stuck :3

9. experimentX

|dw:1362839531405:dw|

10. siddarth95

its all good .. thanks for the responses :)

11. experimentX

there's one another way to do it .. change all trigs into half angles, and change sines and cosines into tan and sec .. you should end up something like|dw:1362841238898:dw|

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