How do I integrate $\dfrac{1}{(\sin x + \cos x)^{4}}$? I could not think of any way. I tried substitution but of no use.

  • $\begingroup$ use the $\sin x + \cos x =\sqrt 2 \sin(\pi/4+x)$ $\endgroup$
    – Math-fun
    Aug 10, 2015 at 19:27
  • $\begingroup$ i would use the tan-half angle substitution $\endgroup$ Aug 10, 2015 at 19:28

2 Answers 2


$$\int \frac{1}{(\sin x + \cos x)^{4}}dx$$


Multiply numerator and denominator by $\sec^4$

$$=\int \frac{\sec^4 (x)} { 1+4\tan(x)+6\tan^2 (x)+4\tan^3(x)+\tan^4(x)}dx$$

Use $\sec^2(x)=\tan^2(x)+1$

$$\int \frac{(1+\tan^2(x))\sec^2(x)}{(1+\tan(x))^4}dx$$

Now substitute $u=\tan(x)$

$$\int \frac{u^2+1}{(u+1)^4}du$$

I hope that you can finish from here.

  • 1
    $\begingroup$ It might be easier to leave the denominator as $(1+\tan x)^4$ and substitute $u = 1+\tan x$. $\endgroup$
    – JimmyK4542
    Aug 10, 2015 at 19:29
  • $\begingroup$ @Nehorai Yes Thanks $\endgroup$
    – Taylor Ted
    Aug 10, 2015 at 19:35

$$\sin x + \cos x = \sqrt{2} \left( \sin x \cos \frac{\pi}{4} + \cos x \sin \frac{\pi}{4} \right) = \sqrt{2} \sin \left( x + \frac{\pi}{4} \right).$$

  • $\begingroup$ what should i do next for integration, next step? $\endgroup$
    – Taylor Ted
    Aug 11, 2015 at 12:21

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.