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Question Number 113275 by bemath last updated on 12/Sep/20

$$\int \frac{\mathrm{dx}}{3\sin \mathrm{x}+\sin {}^3\mathrm{x}}?$$

Answered by bemath last updated on 12/Sep/20

$$\mathrm{I}=\int \frac{\mathrm{dx}}{\sin \mathrm{x}(3+\sin {}^2\mathrm{x})}$$$$\mathrm{I}=\int \frac{\sin \mathrm{x}\mathrm{dx}}{\sin {}^2\mathrm{x}(3+\sin {}^2\mathrm{x})}$$$$\mathrm{I}=\int \frac{-\mathrm{d}\left(\cos \mathrm{x}\right)}{(1-\cos {}^2\mathrm{x})(4-\cos {}^2\mathrm{x})}$$$$\mathrm{I}=\int \frac{-\mathrm{du}}{(1-{\mathrm{u}}^2)(4-{\mathrm{u}}^2)};\mathrm{where}\mathrm{u}=\cos \mathrm{x}$$$$\mathrm{I}=-\int \frac{\mathrm{du}}{(1+\mathrm{u})(1-\mathrm{u})(2+\mathrm{u})(2-\mathrm{u})}$$$$\mathrm{I}=\frac{1}{6}\int \left(\frac{1}{\mathrm{u}-1}\right)-\left(\frac{1}{\mathrm{u}+1}\right)\mathrm{du}-\frac{1}{12}\int (\frac{1}{\mathrm{u}-2}-\frac{1}{\mathrm{u}+2})\mathrm{du}$$$$\mathrm{I}=\frac{1}{6}\ln \mid \frac{\mathrm{u}-1}{\mathrm{u}+1}\mid -\frac{1}{12}\ln \mid \frac{\mathrm{u}-2}{\mathrm{u}+2}\mid +\mathrm{c}$$$$\mathrm{I}=\frac{1}{6}\ln \mid \frac{\cos \mathrm{x}-1}{\cos \mathrm{x}+1}\mid -\frac{1}{12}\ln \mid \frac{\cos \mathrm{x}-2}{\cos \mathrm{x}+2}\mid +\mathrm{c}$$

Answered by abdomsup last updated on 12/Sep/20

$$I=\int \frac{dx}{{sin}^{3}x+3sinx}\Rightarrow$$$$I=\int \frac{dx}{sinx({sin}^{2}x+3)}$$$$letdecomposeF\left(u\right)=\frac{1}{u(u^2+3)}$$$$F\left(u\right)=\frac{a}{u}+\frac{bu+c}{u^2+3}$$$$a=\frac{1}{3},{lim}_{u\to +\mathrm{\infty }}uF\left(u\right)=0$$$$a+b\Rightarrow b=-\frac{1}{3}$$$$F(-u)=-F\left(u\right)\Rightarrow c=0\Rightarrow$$$$F\left(u\right)=\frac{1}{3u}-\frac{u}{3(u^2+3)}$$$$\Rightarrow I=\frac{1}{3}\int \frac{dx}{sinx}-\frac{1}{3}\int \frac{sinx}{{sin}^{2}x+3}$$$$\int \frac{dx}{sinx}{=}_{tsn\left(\frac{x}{2}\right)=t}\int \frac{2dt}{(1+t^2)\frac{2t}{1+t^2}}$$$$=\int \frac{dt}{t}=ln\mid t\mid +c_1=ln\mid tan\left(\frac{x}{2}\right)\mid +c_1$$$$\int \frac{sinx}{{sin}^{2}x+3}dx=\int \frac{sinxdx}{4-{cos}^{2}x}$$$${=}_{cosx=t}\int \frac{-dt}{4-t^2}=\int \frac{dt}{(t-2)(t+2)}$$$$=\frac{1}{4}\int (\frac{1}{t-2}-\frac{1}{t+2})dt=\frac{1}{4}ln\mid \frac{t-2}{t+2}\mid +c_2$$$$=\frac{1}{4}ln\mid \frac{cosx-2}{cosx+2}\mid +c_2\Rightarrow$$$$I=\frac{1}{3}ln\mid tan\left(\frac{x}{2}\right)\mid -\frac{1}{12}ln\left(\frac{2-cosx}{2+cosx}\right)+C$$$$$$

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