Question

Integrate the function w.r.t.x. : $\frac{\sin x}{1+\cos x}$

Answer 1

Here is the answer.

Good luck.

Answer 2

Answer:

$\int\:\frac{sin\;x}{1+cos\:x}dx=-2log(cos\frac{x}{2} )+C$

Step-by-step explanation:

We know that before integrating any function ,we must convert it into integrable form.

As

$sin\:x=2\:sin\:\frac{x}{2} \:cos\frac{x}{2} \\ \\ \\ 1+cos\:x=2\:cos^{2} \frac{x}{2}$

$\int\:\frac{sin\;x}{1+cos\:x}dx\\ \\ \\\int\:\frac{2\:sin\:\frac{x}{2} \:cos\frac{x}{2}}{2\:cos^{2} \frac{x}{2}}$

$=\int\:\frac{sin\frac{x}{2} }{cos\:\frac{x}{2} } dx\\ \\ \\ \int\:\tan\frac{x}{2} dx\\ \\ \\ \int\:\frac{sin\;x}{1+cos\:x}dx=-2log(cos\frac{x}{2} )+C$