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Question Number 36436 by prof Abdo imad last updated on 02/Jun/18

$$find\int \frac{sinx}{1+{cos}^{3}x}dx$$

Answered by tanmay.chaudhury50@gmail.com last updated on 02/Jun/18

$$=\int \frac{-dt}{(1+t)(1-t+t^2)}t=cosx$$$$=\frac{1}{(1+t)(1-t+t^2)}=\frac{a}{1+t}+\frac{bt+c}{1-t+t^2}$$$$1=a-at+{at}^2+bt+c+{bt}^2+ct$$$$$$$$1=a+c+t(-a+b+c)+t^2(a+b)$$$$a+c=1$$$$-a+b+c=0$$$$a+b=0$$$$-a-a+1-a=0$$$$-3a=-1$$$$a=\frac{1}{3}$$$$b=-\frac{1}{3}$$$$\stackrel{}{c}=1-\frac{1}{3}$$$$a=\frac{1}{3}$$$$b=-\frac{1}{3}$$$$c=\frac{2}{3}$$$$=-1\int \frac{\frac{1}{3}}{1+t}dt-\int \frac{\frac{-1}{3}t+\frac{2}{3}}{1-t+t^2}dt$$$$=\frac{-1}{3}ln(1+t)+\frac{1}{3}\int \frac{t-2}{1-t+t^2}dt$$$$=do+\frac{1}{6}\int \frac{2t-4}{t^2-t+1}$$$$=do+\frac{1}{6}\int \frac{2t-1}{t^2-t+1}dt-\frac{1}{6}\int \frac{3}{t^2-t+1}dt$$$$=do+\frac{1}{6}ln(t^2-t+1)-\frac{3}{6}\int \frac{dt}{t^2-2t.\frac{1}{2}+\frac{1}{4}+1-\frac{1}{4}}$$$$=\frac{-1}{3}ln(1+t)+\frac{1}{6}ln(t^2-t+1)-\frac{3}{6}\int \frac{dt}{{(t-\frac{1}{2})}^2+\frac{3}{4}}$$$$do-\frac{1}{2}\times \frac{2}{\sqrt{3}}{tan}^{-1}\left(\frac{t-\frac{1}{2}}{\frac{\sqrt{3}}{2}}\right)$$$$$$

Answered by MJS last updated on 02/Jun/18

$$\int \frac{\sin x}{1+{\cos }^{3}x}dx=$$$$\left[\begin{array}{c}t=2\arctan x\to dx=\frac{2dt}{1+t^2}\\ \sin x=\frac{2t}{1+t^2}\cos x=\frac{1-t^2}{1+t^2}\end{array}\right]$$$$=2\int \frac{t^3+t}{3t^4+1}dt=$$$$[u=t^2\to dt=\frac{du}{2t}]$$$$=\int \frac{u+1}{3u^2+1}du=\int \frac{u}{3u^2+1}du+\int \frac{du}{3u^2+1}=$$$$[v=3u^2+1\to du=\frac{6}{dv};w=\sqrt{3}u\to du=\frac{\sqrt{3}}{3}dw]$$$$=\frac{1}{6}\int \frac{dv}{v}+\frac{\sqrt{3}}{3}\int \frac{dw}{w^2+1}=$$$$=\frac{1}{6}\ln v+\frac{\sqrt{3}}{3}\arctan w=$$$$=\frac{1}{6}\ln (3u^2+1)+\frac{\sqrt{3}}{3}\arctan \left(\sqrt{3}u\right)=$$$$=\frac{1}{6}\ln (3t^4+1)+\frac{\sqrt{3}}{3}\arctan \left(\sqrt{3}{t}^2\right)=$$$$=\frac{1}{6}\ln \left(48(1+{\arctan }^{4}x)\right)+\frac{\sqrt{3}}{3}\arctan \left(4\sqrt{3}{\arctan }^{2}x\right)+C$$

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