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Question Number 42370 by Raj Singh last updated on 24/Aug/18

Commented by maxmathsup by imad last updated on 24/Aug/18

$$letA=\int \frac{dx}{sin\left(2x\right)+sinx}$$$$A=\int \frac{dx}{2sinxcosx+sinx}=\int \frac{sinx}{2{sin}^{2}xcosx+{sin}^{2}x}dx$$$$=\int \frac{sinx}{2(1-{cos}^{2}x)cosx+1-{cos}^{2}x}=\int \frac{sinx}{2cosx-2{cos}^{3}x+1-{cos}^{2}x}dx$$$$=\int \frac{sinxdx}{-2{cos}^{3}x-{cos}^{2}x+2cosx+1}{=}_{cosx=t}\int \frac{-dt}{-2t^3-t^2+2t+1}$$$$=\int \frac{dt}{2t^3+t^2-2t-1}but2t^3+t^2-2t-1=2t^3-2+t^2-2t+1$$$$=2(t-1)(t^2+t+1)+{(t-1)}^2=(t-1)(2t^2+2t+2+t-1)$$$$=(t-1)(2t^2+3t+1)rootsof2t^2+3t+1$$$$\mathrm{\Delta }=9-8=1\Rightarrow t_1=\frac{-3+1}{4}=-\frac{1}{2}and{t}_2=\frac{-3-1}{4}=-1\Rightarrow$$$$2t^3+t^2-2t-1=(t-1)(t+1)(t+\frac{1}{2})letdecompose$$$$F\left(t\right)=\frac{1}{2t^3+t^2-2t-1}=\frac{1}{(t-1)(t+1)(t+\frac{1}{2})}$$$$F\left(t\right)=\frac{a}{t-1}+\frac{b}{t+1}+\frac{c}{t+\frac{1}{2}}$$$$a={lim}_{t\to 1}(t-1)F\left(t\right)=\frac{1}{2.\frac{3}{2}}=\frac{1}{3}$$$$b={lim}_{t\to -1}(t+1)F\left(t\right)=\frac{1}{(-2)(-\frac{1}{2})}=1$$$$c={lim}_{t\to -\frac{1}{2}}(t+\frac{1}{2})F\left(t\right)=\frac{1}{(-\frac{3}{2})\frac{1}{2}}=-\frac{4}{3}\Rightarrow$$$$F\left(t\right)=\frac{1}{3(t-1)}+\frac{1}{t+1}-\frac{4}{3(t+\frac{1}{2})}\Rightarrow$$$$A=\int F\left(t\right)dt=\frac{1}{3}ln\mid t-1\mid +ln\mid t+1\mid -\frac{4}{3}ln\mid t+\frac{1}{2}\mid +c$$$$A=\frac{1}{3}ln\mid cosx-1\mid +ln\mid cosx+1\mid -\frac{4}{3}ln\mid cosx+\frac{1}{2}\mid +c$$$$$$

Answered by tanmay.chaudhury50@gmail.com last updated on 24/Aug/18

$$\int \frac{dx}{sinx(2cosx+1)}$$$$\int \frac{sinxdx}{(1-{cos}^{2}x)(2cosx+1)}$$$$t=cosxdt=-sinxdx$$$$\int \frac{-dt}{(1-t^2)(2t+1)}$$$$\int \frac{dt}{(t+1)(t-1)(2t+1)}$$$$\frac{1}{(t+1)(t-1)(2t+1)}=\frac{a}{t+1}+\frac{b}{t-1}+\frac{c}{2t+1}$$$$1=a(t-1)(2t+1)+b(t+1)(2t+1)+c(t+1)(t-1)$$$$putt-1=0t=1$$$$b(1+1)(2+1)=1\mathit{b}=\frac{1}{6}$$$$\mathit{p}\mathit{u}\mathit{t}\mathit{t}+1=0$$$$\mathit{a}(-1-1)(2\times -1+1)=1$$$$\mathit{a}(-2)(-1)=1\mathit{a}=\frac{1}{2}$$$$\mathit{p}\mathit{u}\mathit{t}2\mathit{t}+1=0$$$$\mathit{c}(\frac{-1}{2}+1)(\frac{-1}{2}-1)=1$$$$\mathit{c}(\frac{1}{4}-1)=1$$$$\mathit{c}\times \frac{-3}{4}=1\mathit{c}=\frac{-4}{3}$$$$\int \frac{a}{t+1}dt+\int \frac{b}{t-1}dt+\int \frac{c}{2t+1}dt$$$$a\int \frac{dt}{t+1}+b\int \frac{dt}{t-1}+\frac{c}{2}\int \frac{dt}{t+\frac{1}{2}}$$$$aln(t+1)+bln(t-1)+\frac{c}{2}ln(t+\frac{1}{2})$$$$\frac{1}{2}ln(cosx+1)+\frac{1}{6}ln(cosx-1)+\frac{-4}{2\times 3}ln(cosx+\frac{1}{2})+k$$$$$$$$$$

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