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.
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$\begingroup$ use the $\sin x + \cos x =\sqrt 2 \sin(\pi/4+x)$ $\endgroup$– Math-funAug 10, 2015 at 19:27
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$\begingroup$ i would use the tan-half angle substitution $\endgroup$– Dr. Sonnhard GraubnerAug 10, 2015 at 19:28
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2 Answers
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$$\int \frac{1}{(\sin x + \cos x)^{4}}dx$$
Hint:
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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1$\begingroup$ It might be easier to leave the denominator as $(1+\tan x)^4$ and substitute $u = 1+\tan x$. $\endgroup$ Aug 10, 2015 at 19:29
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$$\sin x + \cos x = \sqrt{2} \left( \sin x \cos \frac{\pi}{4} + \cos x \sin \frac{\pi}{4} \right) = \sqrt{2} \sin \left( x + \frac{\pi}{4} \right).$$
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$\begingroup$ what should i do next for integration, next step? $\endgroup$ Aug 11, 2015 at 12:21