0
$\begingroup$

How can I evaluate the following integral? $$ \int\frac{\cos{x}}{1+\sin{2x}} dx $$

I tried the following way, but I was not able to proceed further: $$ \begin{gather} I&=\int\frac{\cos{x}}{\left(\sin{x}+\cos{x}\right)^2} dx\\ &= \int\frac{\sec{x}}{\left(1+\tan{x}\right)^2} dx \end{gather} $$

$\endgroup$
  • $\begingroup$ Hope you'll satisty with the solution. $\endgroup$ – Arpit Yadav Aug 12 '18 at 3:01
2
$\begingroup$

Hint:

Write numerator as $$2\cos x=\cos x+\sin x +(\cos x-\sin x)$$

$\endgroup$
0
$\begingroup$

Using lab bhattacharjee result,$$I=\int \frac{\cos x+ \sin x}{{(\cos x+ \sin x})^2}dx\space +\int \frac{\cos x-\sin x}{(\cos x+\sin x)^2}dx$$ $$I=I_1+I_2$$ $$I_1=\int \frac{1}{\cos x+\sin x}dx$$ $$I_1=\int\frac{1}{\sqrt2(\cos x\cdot\cos{\frac{\pi}{4}+\sin x\cdot \sin{\frac{\pi}{4}})}}dx$$ $$I_1=\int\frac{\sec(x-\pi/4)}{\sqrt2}dx$$ $$I_1=\frac{1}{\sqrt2}\cdot\ln[(\sec(x-\pi/4)+\tan(x-\pi/4)]+C_1$$ Now$$I_2=\int \frac{\cos x-\sin x}{(\cos x+\sin x)^2}dx$$ $$\sin x+\cos x=t\implies(\cos x-\sin x)dx=dt$$ $$I_2=\int\frac {1}{t^2}\space dt$$ $$I_2=-\frac{1}{t}+C_2=-\frac{1}{\sin x+\cos x}+C_2$$

$\endgroup$

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.