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Question Number 174628 by cortano1 last updated on 06/Aug/22

$$\mathrm{\Omega }=\int \frac{x}{1+\csc x}dx=?$$

Commented by infinityaction last updated on 06/Aug/22

$$\int \frac{x}{1+\csc x}dx=\int \frac{xs\mathrm{in}x}{1+\sin x}dx$$$$=\int x\frac{1+\sin x-1}{1+\sin x}dx=\int (x-\frac{x}{1+\sin x})dx$$$$\mathrm{\Omega }=\frac{x^2}{2}-\int \frac{x}{\underset{I}{\underbrace{1+\sin x}}}$$$$I=\int \frac{x}{1+\sin x}\Rightarrow I=\int \frac{x}{1+\cos (\frac{\pi }{2}-x)}$$$$I=\int \frac{x}{{2\cos }^2(\frac{\pi }{4}-\frac{x}{2})}$$$$\frac{\pi }{4}-\frac{x}{2}=y\Rightarrow \frac{dx}{2}=-dy$$$$I=-\int \frac{(\frac{\pi }{2}-2y)dy}{{\cos }^{2}y}$$$$I=2\int y{\sec }^{2}ydy-\frac{\pi }{2}\int {\sec }^{2}ydy$$$$I=2y\tan y-\log \mid \sin y\mid -\frac{\pi }{2}\tan y+C$$$$\mathrm{\Omega }=\frac{x^2}{2}+\log \mid \sin y\mid +\frac{\pi }{2}\tan y-2y\tan y+C$$$$herey=\frac{\pi }{4}-\frac{x}{2}$$

Commented by Tawa11 last updated on 09/Aug/22

$$\mathrm{Great}\mathrm{sir}$$

Answered by aleks041103 last updated on 06/Aug/22

$$\int \frac{x}{1+cscx}dx=\int \frac{xsinx}{1+sinx}dx=$$$$=\int x\frac{1+sinx-1}{1+sinx}dx=\int (x-\frac{x}{1+sinx})dx=$$$$=\frac{x^2}{2}-\int \frac{xdx}{1+sinx}=\frac{x^2}{2}+I$$$$z=tan\left(x/2\right)\Rightarrow x=2arctan\left(z\right)\Rightarrow dx=\frac{2dz}{1+z^2}$$$$sin\left(x\right)=2sin\left(x/2\right)cos\left(x/2\right)=$$$$=2tan\left(x/2\right){cos}^2\left(x/2\right)=$$$$=\frac{2tan\left(x/2\right)}{1+{tan}^2\left(x/2\right)}=\frac{2z}{1+z^2}$$$$\Rightarrow \frac{1}{1+sinx}=\frac{1+z^2}{{(1+z)}^2}$$$$\Rightarrow I=4\int arctan\left(z\right)(-\frac{dz}{{(1+z)}^2})=$$$$=4\int arctan\left(z\right)d\left(\frac{1}{1+z}\right)=$$$$=\frac{4arctan\left(z\right)}{1+z}-4\int \frac{dz}{(1+z)(1+z^2)}=$$$$=\frac{2x}{1+tan\left(x/2\right)}+2\int \frac{-2dz}{(1+z)(1+z^2)}$$$$\frac{-2}{(1+z)(1+z^2)}=\frac{a}{1+z}+\frac{bz+c}{1+z^2}$$$$\Rightarrow a+{az}^2+bz+c+{bz}^2+cz=-2$$$$a+c=-2,b+c=0,a+b=0\Rightarrow -a=-c=b=1$$$$\Rightarrow \int \frac{-2dz}{(1+z)(1+z^2)}=\int (\frac{z-1}{1+z^2}-\frac{1}{1+z})dz=$$$$=\int \frac{zdz}{1+z^2}-\int \frac{dz}{1+z^2}-\int \frac{dz}{1+z}=$$$$=\frac{1}{2}\int \frac{d(1+z^2)}{1+z^2}-arctan\left(z\right)-ln(1+z)=$$$$=\frac{1}{2}ln(1+z^2)-ln(1+z)-arctan\left(z\right)+C=$$$$=\frac{1}{2}ln\left(\frac{1+z^2}{{(1+z)}^2}\right)-arctan\left(z\right)+C=$$$$=-\frac{1}{2}ln(1+sinx)-\frac{1}{2}x+C$$$$\Rightarrow I=\frac{2x}{1+tan\left(x/2\right)}-x-ln(1+sinx)+C$$$$\Rightarrow \int \frac{x}{1+cscx}dx=\frac{x^2}{2}+\frac{2x}{1+tan\left(x/2\right)}-x-ln(1+sinx)+C$$

Commented by Tawa11 last updated on 09/Aug/22

$$\mathrm{Great}\mathrm{sir}$$

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