integral of 1/(sinx+cosx): u-substitution??

[thebenji]

int[1/(sinx+cosx) dx

i really have no idea how to begin this. it isn't even a homework problem--how would i start? i can't think of an obvious u-substitution

 

[pka]

You are not going to find an elementary closed form.
It can be done in the complex number field.
This web site will give an answer.
http://integrals.wolfram.com/index.jsp

 

[galactus]

This problem can be done with elementary means, but it's a booger.

If you use the Weierstrauss substitutions along with some other choice substitutions, you can whittle away at it.

\(\displaystyle \L\\\int\frac{1}{sin(x)+cos(x)}dx\)

Use \(\displaystyle \L\\u=tan(\frac{x}{2}); \;\ sin(x)=\frac{2u}{1+u^{2}}; \;\ cos(x)=\frac{1-u^{2}}{1+u^{2}}; \;\ dx=\frac{2}{1+u^{2}}du\)

If you sub these in, you get:

\(\displaystyle \L\\-2\int\frac{1}{u^{2}-2u-1}du\)

This is still tough to integrate, so make another substitution, maybe

\(\displaystyle \L\\u=w+1; \;\ du=dw\)

This gives:

\(\displaystyle \L\\-2\int\frac{1}{w^{2}-2}dw=\frac{-ln(\frac{\sqrt{2}w-2}{\sqrt{2}w+2})}{\sqrt{2}}\)

Resub w=u-1:

\(\displaystyle \L\\\frac{-ln(\frac{\sqrt{2}u-\sqrt{2}-2}{\sqrt{2}u-\sqrt{2}+2})}{\sqrt{2}}\)

Now, Resub u=tan(x/2)

 

[tkhunny]

It's a little easier to start with this, isn't it?

\(\displaystyle \L\,\frac{1}{sin(x)+cos(x)} = \frac{1}{\sqrt{2}sin(x+\frac{\pi}{4})}\)