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Question Number 97782 by 175 last updated on 09/Jun/20

$$Evaluate:$$$$\int \frac{sinx}{1+{sin}^{2}x}dx$$

Commented by mr W last updated on 09/Jun/20

$$=-\int \frac{d\left(\cos x\right)}{(\sqrt{2}-\cos x)(\sqrt{2}+\cos x)}$$$$=-\frac{\sqrt{2}}{4}\int [\frac{1}{\sqrt{2}-\cos x}+\frac{1}{\sqrt{2}+\cos x}]d\left(\cos x\right)$$$$=\frac{\sqrt{2}}{4}\ln \frac{\sqrt{2}-\cos x}{\sqrt{2}+\cos x}+C$$

Answered by niroj last updated on 09/Jun/20

$$\int \frac{\sin \mathrm{x}}{1+{\sin }^2\mathrm{x}}\mathrm{dx}$$$$\int \frac{\sin \mathrm{x}\mathrm{dx}}{1+1-{\cos }^2\mathrm{x}}=\int \frac{\sin \mathrm{x}\mathrm{dx}}{2-{\cos }^2\mathrm{x}}$$$$\mathrm{Put},\cos \mathrm{x}=\mathrm{t}$$$$-\sin \mathrm{x}\mathrm{dx}=\mathrm{dt}$$$$\mathrm{sinxdx}=-\mathrm{dt}$$$$-\int \frac{1}{2-{\mathrm{t}}^2}\mathrm{dt}$$$$=-\int \frac{1}{{\left(\sqrt{2}\right)}^2-{\left(\mathrm{t}\right)}^2}\mathrm{dt}$$$$=-\frac{1}{2\sqrt{2}}\log \frac{\sqrt{2}+\mathrm{t}}{\sqrt{2}-\mathrm{t}}+\mathrm{C}$$$$=-\frac{1}{2\sqrt{2}}\log \frac{\sqrt{2}+\cos \mathrm{x}}{\sqrt{2}-\cos \mathrm{x}}+\mathrm{C}//.$$$$$$

Answered by abdomathmax last updated on 09/Jun/20

$$\mathrm{I}=\int \frac{\mathrm{sinx}\mathrm{dx}}{1+{\sin }^2\mathrm{x}}\Rightarrow \mathrm{I}=\int \frac{\mathrm{sinxdx}}{2-{\cos }^2\mathrm{x}}$$$$\mathrm{we}\mathrm{do}\mathrm{the}\mathrm{changement}\mathrm{cosx}=\mathrm{t}\Rightarrow$$$$\mathrm{I}=\int \frac{-\mathrm{dt}}{2-{\mathrm{t}}^2}=\int \frac{\mathrm{dt}}{{\mathrm{t}}^2-2}=\frac{1}{2\sqrt{2}}\int (\frac{1}{\mathrm{t}-\sqrt{2}}-\frac{1}{\mathrm{t}+\sqrt{2}})\mathrm{dt}$$$$=\frac{1}{2\sqrt{2}}\ln \mid \frac{\mathrm{t}-\sqrt{2}}{\mathrm{t}+\sqrt{2}}\mid +\mathrm{C}\Rightarrow$$$$\mathrm{I}=\frac{1}{2\sqrt{2}}\ln \mid \frac{\mathrm{cosx}-\sqrt{2}}{\mathrm{cosx}+\sqrt{2}}\mid +\mathrm{C}$$

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