Can these integrals be solved by u-substitution using trig identities?

Question

$\int \frac{\operatorname{sin} x}{2\sin x\cos x+5}dx$

$\int \frac{\operatorname{sin}x\cos x}{\operatorname{sin}x+\cos x}dx$

$\int \frac{dx}{\operatorname{sin} x+\cos x}$

I'd like to know if there's a way to solve some of these integrals by manipulating it and then doing u substitution, I tried but I'm not very good at trig identities. I got the second one like this:

$\int \frac{\operatorname{sin}2x(\operatorname{sin}x-\cos x)}{\cos 2x}dx$ but didn't know how to go from there.

Answer 1

Assuming $\operatorname{sen}$ is synonymous with $\sin$, the third integral is $$\int\tfrac{1}{\sqrt{2}}\csc (x+\tfrac{\pi}{4})dx=-\tfrac{1}{\sqrt{2}}\ln |\csc (x+\tfrac{\pi}{4})+\cot (x+\tfrac{\pi}{4})|+C=-\tfrac{1}{\sqrt{2}}\ln |\cot (\tfrac{x}{2}+\tfrac{\pi}{8})|+C,$$while the first integral is very messy (although integrating it between certain limits might be neater). Just to expand on lab bhattacharjee's treatment of the second integral,$$\int\frac{1}{\sqrt{8}}\frac{\sin (2y-\tfrac{\pi}{2})}{\sin y}dy=-\int\frac{1}{\sqrt{8}}\frac{\cos 2y}{\sin y}dy=\int\frac{2\sin y-\csc y}{\sqrt{8}}dy\\=\frac{-2\cos y+\ln|\csc y+\cot y|}{\sqrt{8}}+C=\frac{-2\cos (x+\tfrac{\pi}{4})+\ln|\cot (\tfrac{x}{2}+\tfrac{\pi}{8})|}{\sqrt{8}}+C.$$

Answer 2

Hint:

$$\dfrac{\sin x\cos x}{\sin x+\cos x}=\dfrac{\sin2x}{2\sqrt2\sin(x+\pi/4)}$$

Set $x+\pi/4=y$

$\cos2y=1-2\sin^2y$

Answer 3

For the first one, write numerator as $$\dfrac{2\sin x}{2\sin x\cos x+5}=\dfrac{\sin x-\cos x}{2(2\sin x\cos x+5)}+\dfrac{\sin x+\cos x}{2(2\sin x\cos x+5)}$$

For the first part, write $$\int(\sin x-\cos x)dx=u, u^2=1+2\sin x\cos x\iff\sin x\cos x=?$$

For the second, $$\int(\sin x+\cos x)dx=v, v^2=1-2\sin x\cos x\iff\sin x\cos x=?$$$