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Question Number 154195 by rexford last updated on 15/Sep/21

$$\int \frac{x+sinx}{1+cosx}dx$$$$please,helpme$$

Answered by qaz last updated on 15/Sep/21

$$\because {\left(\frac{\sin \mathrm{x}}{1+\cos \mathrm{x}}\right)}^{\prime }=\frac{1}{1+\cos \mathrm{x}}$$$$\therefore \int \frac{\mathrm{x}+\sin \mathrm{x}}{1+\cos \mathrm{x}}\mathrm{dx}$$$$=\int \frac{\mathrm{x}}{1+\cos \mathrm{x}}\mathrm{dx}+\int \frac{\sin \mathrm{x}}{1+\cos \mathrm{x}}\mathrm{dx}$$$$=\int \mathrm{xd}\left(\frac{\sin \mathrm{x}}{1+\cos \mathrm{x}}\right)+\int \frac{\sin \mathrm{x}}{1+\cos \mathrm{x}}\mathrm{dx}$$$$=\frac{\mathrm{xsin}\mathrm{x}}{1+\cos \mathrm{x}}+\mathrm{C}$$

Commented by qaz last updated on 15/Sep/21

$$\int \frac{\mathrm{x}-\sin \mathrm{x}}{1+\cos \mathrm{x}}\mathrm{dx}......\int \frac{\mathrm{x}-\sin \mathrm{x}}{1-\cos \mathrm{x}}\mathrm{dx}......\int \frac{\mathrm{x}+\cos \mathrm{x}}{1+\sin \mathrm{x}}\mathrm{dx}.....\int \frac{\mathrm{x}-\cos \mathrm{x}}{1+\sin \mathrm{x}}\mathrm{dx}$$$$\int \frac{\mathrm{x}+\cos \mathrm{x}}{1-\sin \mathrm{x}}\mathrm{dx}......\int \frac{\mathrm{x}-\cos \mathrm{x}}{1-\sin \mathrm{x}}\mathrm{dx}.......\int \frac{\mathrm{x}+\sin \mathrm{x}}{1-\cos \mathrm{x}}\mathrm{dx}$$$$\mathrm{The}\mathrm{same}....$$

Answered by puissant last updated on 15/Sep/21

$$\mathrm{\Omega }=\int \frac{x+sinx}{1+cosx}dx$$$$=\int \frac{x}{1+cosx}dx+\int \frac{sinx}{1+cosx}dx$$$$=\int {xsec}^2\left(\frac{x}{2}\right)dx-ln\mid cosx\mid +C$$$$=xtan\left(\frac{x}{2}\right)-\int tan\left(\frac{x}{2}\right)dx-ln\mid cosx\mid +C$$$$=xtan\left(\frac{x}{2}\right)+2ln\mid cos\left(\frac{x}{2}\right)\mid -ln\mid cosx\mid +C$$$$$$$$\therefore \because \mathrm{\Omega }=xtan\left(\frac{x}{2}\right)+2ln\mid cos\left(\frac{x}{2}\right)\mid -ln\mid cosx\mid +C..$$

Answered by maged last updated on 15/Sep/21

$$I=\int \frac{x}{1+\cos x}.\frac{1-\cos x}{1-\cos x}dx+\int \frac{\sin x}{1+\cos x}.\frac{1-\cos x}{1-\cos x}dx$$$$=\int \frac{x(1-\cos x)}{{\sin }^{2}x}dx+\int \frac{1}{\sin x}dx$$$$I=\int x{\mathrm{cosec}}^{2}xdx-\int x\cot x\mathrm{cosec}xdx+\int \mathrm{cosec}xdx$$$$u=x\to du=dx$$$$dv={\mathrm{cosec}}^{2}xdx\to v=-\cot x$$$$--$$$$u=x\to du=dx$$$$dv=-\cot x\mathrm{cosec}xdx\to v=\mathrm{cosec}x$$$$--$$$$I=-x\cot x+x\mathrm{cosec}x+\int \cot xdx-\int \mathrm{cosec}xdx+\int \mathrm{cosec}xdx$$$$=x\mathrm{cosec}x-x\cot x-\ln \left(\sin x\right)+C$$

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