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Question Number 118419 by bramlexs22 last updated on 17/Oct/20

$$\int \frac{2\sin 2x}{4\cos x+\sin 2x}dx$$

Commented by bramlexs22 last updated on 17/Oct/20

$$yes...thankyouallmaster$$

Answered by peter frank last updated on 17/Oct/20

$$\int \frac{4\sin x\cos x}{4\cos x+2\sin x\cos x}dx$$$$\int \frac{2\sin x}{2+\sin x}$$$$.....$$

Commented by TANMAY PANACEA last updated on 17/Oct/20

$$sinx=\frac{2t}{1+t^2}t=tan\frac{x}{2}\to 2dt={sec}^2\frac{x}{2}dt$$$$\int \frac{2\times \frac{2t}{1+t^2}\times \frac{2dt}{1+t^2}}{2+\frac{2t}{1+t^2}}$$$$\int \frac{8tdt}{{(1+t^2)}^2}\times \frac{1+t^2}{2+2t^2+2t}$$$$\int \frac{4t(1+t^2)}{{(1+t^2)}^2(1+t+t^2)}dt$$$$4\int \frac{tdt}{(1+t^2)(1+t+t^2)}$$$$4\int \frac{(1+t+t^2)-(1+t^2)}{(1+t^2)(1+t+t^2)}dt$$$$4\int \frac{dt}{1+t^2}-4\int \frac{dt}{t^2+2\times t\times \frac{1}{2}+\frac{1}{4}+\frac{3}{4}}$$$$4\int \frac{dt}{1+t^2}-4\int \frac{dt}{{(t+\frac{1}{2})}^2+{\left(\frac{\sqrt{3}}{2}\right)}^2}$$$$4{tan}^{-1}\left(t\right)-4\times \frac{1}{\left(\frac{\sqrt{3}}{2}\right)}{tan}^{-1}\left(\frac{t+\frac{1}{2}}{\frac{\sqrt{3}}{2}}\right)+c$$$$putt=tan\frac{x}{2}$$$$$$

Commented by peter frank last updated on 18/Oct/20

$$\mathrm{thank}\mathrm{you}\mathrm{all}$$

Answered by Dwaipayan Shikari last updated on 17/Oct/20

$$2\int \frac{sinx}{2+sinx}dx$$$$2\int 1-\frac{2}{2+sinx}=2x-8\int \frac{1}{2+\frac{2t}{1+t^2}}.\frac{1}{1+t^2}dt(t=tan\frac{x}{2})$$$$=2x-4\int \frac{1}{1+t^2+t}dt$$$$=2x-4\int \frac{1}{{(t+\frac{1}{2})}^2+{\left(\frac{\sqrt{3}}{2}\right)}^2}dt$$$$=2x-\frac{8}{\sqrt{3}}{tan}^{-1}\frac{2t+1}{\sqrt{3}}+C$$$$=2x-\frac{8}{\sqrt{3}}{tan}^{-1}\frac{2tan\frac{x}{2}+1}{\sqrt{3}}+C$$

Answered by benjo_mathlover last updated on 17/Oct/20

$$solve\int \frac{2\sin 2x}{4\cos x+\sin 2x}dx.$$$$Solution:$$$$I=\int \frac{4\sin x\cos x}{4\cos x+2\sin x\cos x}dx=\int \frac{2\sin x\cos x}{2\cos x+\sin x\cos x}dx$$$$I=\int \frac{2\sin x}{2+\sin x}dx;let\tan \left(\frac{x}{2}\right)=z\Rightarrow \frac{1}{2}\sec {}^2\left(\frac{x}{2}\right)dx=dz$$$$ordx=2\cos {}^2\left(\frac{x}{2}\right)dz$$$$I=2\int \left[\frac{\frac{2z}{1+z^2}}{\frac{2z^2+2z+2}{1+z^2}}\right]\left(\frac{2}{1+z^2}\right)dz=2\int \left[\frac{2z}{(z^2+z+1)(z^2+1)}\right]dz$$$$\frac{2z}{(z^2+z+1)(z^2+1)}=\frac{az+b}{z^2+z+1}+\frac{cz+d}{z^2+1}$$$$\Rightarrow 2z=(z^2+1)(az+b)+(cz+d)(z^2+z+1)$$$$z=0\Rightarrow 0=b+d;d=-b$$$$z=1\Rightarrow 2=2a+2b+3c+3d;2=2a-b+3c$$$$z=-1\Rightarrow -2=-2a+2b-c+d;-2=-2a+b-c$$$$z=-2\Rightarrow -4=-10a+5b-6c+3d;-2=-5a+b-3c$$$$a=0;b=-2;c=0;d=2$$$$I=2[-2\int \frac{dz}{z^2+z+1}+2\int \frac{dz}{z^2+1}]$$$$I=4{\tan }^{-1}\left(z\right)-4\int \frac{dz}{{(z+\frac{1}{2})}^2+{\left(\frac{\sqrt{3}}{2}\right)}^2}$$$$secondintegralletz+\frac{1}{2}=\frac{\sqrt{3}}{2}\tan r\Rightarrow r={\tan }^{-1}\left(\frac{2z+1}{\sqrt{3}}\right)$$$$I_2=-4.\frac{\sqrt{3}}{2}\int \frac{\sec {}^{2}rdr}{\frac{3}{4}\sec {}^{2}r}=-\frac{8}{\sqrt{3}}r+c$$$$I_2=-\frac{8}{\sqrt{3}}{\tan }^{-1}\left(\frac{2z+1}{3}\right)+c$$$$ThusI={4\tan }^{-1}\left(z\right)-\frac{8}{\sqrt{3}}{\tan }^{-1}\left(\frac{2z+1}{3}\right)+c$$$$I={4\tan }^{-1}\left(\tan \left(\frac{x}{2}\right)\right)-\frac{8}{\sqrt{3}}{\tan }^{-1}\left(\frac{2\tan \left(\frac{x}{2}\right)+1}{3}\right)+c$$$$orI=2x-\frac{8}{\sqrt{3}}{\tan }^{-1}\left(\frac{2\tan \left(\frac{x}{2}\right)+1}{\sqrt{3}}\right)+c$$$$$$

Answered by 1549442205PVT last updated on 17/Oct/20

$$\int \frac{2\sin 2x}{4\cos x+\sin 2x}dx=\int \frac{4\mathrm{sinxcosxdx}}{2\mathrm{cosx}(2+\mathrm{sinx})}$$$$=\int \frac{2\mathrm{sinxdx}}{2+\mathrm{sinx}}=\int (2-\frac{4}{2+\mathrm{sinx}})\mathrm{dx}$$$$=2\mathrm{x}-4\int \frac{\mathrm{dx}}{2+\mathrm{sinx}}.\left(1\right)\mathrm{Put}\tan \frac{\mathrm{x}}{2}=\mathrm{t}$$$$\Rightarrow \mathrm{dt}=\frac{1}{2}(1+{\mathrm{t}}^2)\mathrm{dx}$$$$\int \frac{\mathrm{dx}}{2+\mathrm{sinx}}=\int \frac{2\mathrm{dt}}{(1+{\mathrm{t}}^2)(2+\frac{2\mathrm{t}}{1+{\mathrm{t}}^2})}$$$$=\int \frac{2\mathrm{dt}}{{2\mathrm{t}}^2+2\mathrm{t}+2}=\int \frac{\mathrm{dt}}{{\mathrm{t}}^2+\mathrm{t}+1}=\int \frac{\mathrm{dt}}{{(\mathrm{t}+\frac{1}{2})}^2+{\left(\frac{\sqrt{3}}{2}\right)}^2}$$$$=\frac{2}{\sqrt{3}}{\tan }^{-}\left(\frac{2\mathrm{t}+1}{\sqrt{3}}\right)\left(2\right).\mathrm{Since}\int \frac{\mathrm{dx}}{{\mathrm{x}}^2+{\mathrm{a}}^2}=\frac{1}{\mathrm{a}}{\tan }^{-1}\left(\frac{\mathrm{x}}{\mathrm{a}}\right)$$$$\mathrm{From}\left(1\right)\left(2\right)\mathrm{we}\mathrm{get}:$$$$\int \frac{2\sin 2x}{4\cos x+\sin 2x}dx$$$$=2\mathrm{x}-\frac{8}{\sqrt{3}}{\tan }^{-1}\left(\frac{2\tan \frac{\mathrm{x}}{2}+1}{\sqrt{3}}\right)+\mathrm{C}$$

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