0
$\begingroup$

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

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

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

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.

$\endgroup$
1
$\begingroup$

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,$$\int\frac{1}{\sqrt{8}}\frac{\sin (2y-\tfrac{\pi}{2})}{\sin y}dy=-\int\frac{1}{\sqrt{8}}\frac{\cos 2y}{\sin y}dy=\int\frac{2\sin y-\csc y}{\sqrt{8}}dy\\=\frac{-2\cos y+\ln|\csc y+\cot y|}{\sqrt{8}}+C=\frac{-2\cos (x+\tfrac{\pi}{4})+\ln|\cot (\tfrac{x}{2}+\tfrac{\pi}{8})|}{\sqrt{8}}+C.$$

$\endgroup$
2
$\begingroup$

Hint:

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

Set $x+\pi/4=y$

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

$\endgroup$
2
$\begingroup$

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

$\endgroup$
  • $\begingroup$ Could you expand a Little bit? $\endgroup$ – Andres Oropeza Nov 24 '18 at 19:21
  • $\begingroup$ @AndresOropeza, Please find the updated answer $\endgroup$ – lab bhattacharjee Nov 25 '18 at 16:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.