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 '15 at 19:27
  • $\begingroup$ i would use the tan-half angle substitution $\endgroup$ – Dr. Sonnhard Graubner Aug 10 '15 at 19:28

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


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.

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    $\begingroup$ It might be easier to leave the denominator as $(1+\tan x)^4$ and substitute $u = 1+\tan x$. $\endgroup$ – JimmyK4542 Aug 10 '15 at 19:29
  • $\begingroup$ @Nehorai Yes Thanks $\endgroup$ – Taylor Ted Aug 10 '15 at 19:35

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

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  • $\begingroup$ what should i do next for integration, next step? $\endgroup$ – Taylor Ted Aug 11 '15 at 12:21

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