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This question already has an answer here:

One way that I tried to solve an integral was to divide integrand by $\cos^3x$ para para obter $$ \int_{0}^{\pi/2}\dfrac{\sec^3x}{1 + \tan^3x}dx $$ Through software, I saw that the result involves $\tanh^{-1}\bigl(\frac{1}{\sqrt{2}}\bigr)$, but I do not know how to get there.

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marked as duplicate by Nosrati, user91500, Arnaud D., Namaste integration Aug 27 '18 at 11:15

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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If $\sin x+\cos x=t,t^2=?$

$$\dfrac2{t(3-t^2)}=\dfrac23\dfrac{t^2+3-t^2}{t(3-t^2)}=?$$

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  • $\begingroup$ Thank you. Good Job. $\endgroup$ – Mathsource Aug 27 '18 at 5:57
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Let $\tan\frac{x}{2}=t$.

Thus, $$\int\limits_0^{\frac{\pi}{2}}\frac{1}{\sin^3x+\cos^3x}dx=\int\limits_0^1\frac{2}{\left(\left(\frac{2t}{1+t^2}\right)^3+\left(\frac{1-t^2}{1+t^2)}\right)^3\right)(1+t^2)}dt=$$ $$=2\int\limits_0^1\frac{(1+t^2)^2}{(2t+1-t^2)((2t)^2-2t(1-t^2)+(1-t^2)^2)}dt.$$ Now, $$2t+1-t^2=-(t^2-2t+1-2)=-(t-1-\sqrt2)(t-1+\sqrt2)$$ and $$(2t)^2-2t(1-t^2)+(1-t^2)^2)=t^4+2t^3+2t^2-2t+1=$$ $$=(t^2+t+k)^2-((2k-1)t^2+2(k+1)t+k^2-1)=$$ $$=(t^2+t+2)^2-((2\cdot2-1)t^2+2(2+1)t+2^2-1)=$$ $$=(t^2+t+2)^2-3(t+1)^2=$$ $$=(t^2-(\sqrt3-1)t+2-\sqrt3)(t^2+(1+\sqrt3)t+2+\sqrt3).$$ Can you end it now?

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