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Question Number 129177 by bramlexs22 last updated on 13/Jan/21

$$\mathrm{G}=\int \frac{\sin {}^2\mathrm{x}+\sin \mathrm{x}}{1+\sin \mathrm{x}+\cos \mathrm{x}}\mathrm{dx}?$$

Answered by liberty last updated on 13/Jan/21

$$\mathrm{G}=\int \frac{\sin \mathrm{x}(\sin \mathrm{x}+1)}{\sin \mathrm{x}+2\cos {}^2\left(\frac{\mathrm{x}}{2}\right)}\mathrm{dx}$$$$\mathrm{G}=\int \frac{2\sin \left(\frac{\mathrm{x}}{2}\right)\cos \left(\frac{\mathrm{x}}{2}\right){\{\sin \left(\frac{\mathrm{x}}{2}\right)+\cos \left(\frac{\mathrm{x}}{2}\right)\}}^2}{2\cos \left(\frac{\mathrm{x}}{2}\right)\{\sin \left(\frac{\mathrm{x}}{2}\right)+\cos \left(\frac{\mathrm{x}}{2}\right)\}}\mathrm{dx}$$$$\mathrm{G}=\int 2\sin \left(\frac{\mathrm{x}}{2}\right)\{\sin \left(\frac{\mathrm{x}}{2}\right)+\cos \left(\frac{\mathrm{x}}{2}\right)\}\mathrm{dx}$$$$\mathrm{G}=\int (1-\cos \mathrm{x}+\sin \mathrm{x})\mathrm{dx}=\underset{―}{\mathrm{x}-\sin \mathrm{x}-\cos \mathrm{x}+\mathrm{c}}$$$$$$

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