Question

Extreme challenge: how to integrate sinx/(sinx+cosx)

Answer 1

Well, use the quotient rule here.

Answer 2

if i recall correctly the gimmick is to multiply top and bottom by the conjugate of the denominator, which gives you something easier to integrate

Answer 3

@satellite73 the expression becomes a mountain after using conjugates, anything simpler?

Answer 4

lol did anyone try?

Answer 5

I take it back, I don't see any other way to do this than with the Weierstrauss sub.
Let
\[u=\tan\left(\frac{x}{2}\right)\\
du=\frac{1}{2}\sec^2\left(\frac{x}{2}\right)\;dx\\
2\cos^2\left(\frac{x}{2}\right)\;du=dx \;\;\;\;\;(*)\]
So, using this sub you have the following situation:
And from this reference triangle, you get
\[\cos^2\left(\frac{x}{2}\right)=\frac{1}{u^2+1},\]
which you substitute in for (*):
\[\frac{2}{u^2+1}du=dx\]
Now, using some identities, you have
\[\begin{align*}\sin x &= \sin\left(2\frac{x}{2}\right)\\
&=2\sin\left(\frac{x}{2}\right)\cos\left(\frac{x}{2}\right)\\
&=2\left(\frac{u}{\sqrt{u^2+1}}\right)\left(\frac{1}{\sqrt{u^2+1}}\right)\\
&=\frac{2u^2}{u^2+1}\end{align*}\]
Likewise, you have
\[\begin{align*}\cos x &= \cos\left(2\frac{x}{2}\right)\\
&=\cos^2\left(\frac{x}{2}\right) - \sin^2\left(\frac{x}{2}\right)\\
&=\left(\frac{1}{\sqrt{u^2+1}}\right)^2-\left(\frac{u}{\sqrt{u^2+1}}\right)^2\\
&=\frac{1-u^2}{u^2+1}\end{align*}\]
So, after all this work, the integral changes to
\[\int \frac{\frac{2u^2}{u^2+1}}{\frac{2u^2}{u^2+1}+\frac{1-u^2}{u^2+1}}\cdot \frac{2}{u^2+1}du\]

Answer 6

@sithsandgiggles what did you just do i don't understand

Answer 7

It's called the Weierstrauss substitution: http://planetmath.org/WeierstrassSubstitutionFormulas.html
I just derived it in its entirety.

Answer 8

Painstakingly, I might add... :)

Answer 9

It's a quirky substitution that ALWAYS works. Generally, if you use it, it is very likely you overlooked something cleaner! Not always, just VERY LIKELY.