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Question Number 94739 by 174 last updated on 20/May/20

Commented by i jagooll last updated on 20/May/20

$$\left(1\right)\int \frac{\cos \mathrm{x}\mathrm{dx}}{3\sin \mathrm{x}-2(1-\sin {}^2\mathrm{x})}=$$$$\int \frac{\mathrm{d}\left(\sin \mathrm{x}\right)}{2\sin {}^2\mathrm{x}+3\sin \mathrm{x}-2}=\int \frac{\mathrm{du}}{(2\mathrm{u}-1)(\mathrm{u}+2)}$$$$\mathrm{now}\mathrm{it}\mathrm{easy}\mathrm{to}\mathrm{solve}$$

Commented by PRITHWISH SEN 2 last updated on 20/May/20

$$3)\mathrm{put}\mathrm{sinx}=\sqrt{2}\mathrm{t}\Rightarrow \mathrm{cosxdx}=\sqrt{2}\mathrm{dt}$$$$\int \frac{\mathrm{dt}}{\sqrt{2}(1+{\mathrm{t}}^2)}=\frac{1}{\sqrt{2}}{\tan }^{-1}\left(\frac{\sin \mathrm{x}}{\sqrt{2}}\right)+\mathrm{C}$$$$4)\int {\sec }^2\mathrm{xdx}+\int {\tan }^2\mathrm{x}\mathrm{dx}=2\int \sec {}^2\mathrm{x}\mathrm{dx}-\int \mathrm{dx}$$$$=2\tan \mathrm{x}-\mathrm{x}+\mathrm{C}$$$$5)\int [\mathrm{x}+\ln (\mathrm{e}-\mathrm{x})]\mathrm{dx}=\frac{{\mathrm{x}}^2}{2}+\mathrm{xln}(\mathrm{e}-\mathrm{x})+\int \frac{\mathrm{x}}{\mathrm{e}-\mathrm{x}}\mathrm{dx}$$$$=\frac{{\mathrm{x}}^2}{2}+\mathrm{xln}(\mathrm{e}-\mathrm{x})-\mathrm{eln}(\mathrm{e}-\mathrm{x})-\mathrm{x}+\mathrm{C}$$

Answered by Rio Michael last updated on 20/May/20

$$\int \frac{dx}{e^{-x}+e^x}=\int \frac{e^x}{1+e^{2x}}dx=\int \frac{1}{1+e^{2x}}{e}^{x}dx$$$$\mathrm{let}u=e^x\Rightarrow du=e^{x}dx$$$$\Rightarrow \int \frac{1}{1+e^{2x}}{e}^{x}dx=\int \frac{1}{1+u^2}du={\tan }^{-1}u+k={\tan }^{-1}{e}^x+k$$

Answered by i jagooll last updated on 20/May/20

$$\left(2\right)\int \frac{{\mathrm{e}}^{\mathrm{x}}}{1+{\mathrm{e}}^{2\mathrm{x}}}\mathrm{dx}=\int \frac{\mathrm{du}}{1+{\mathrm{u}}^2},[\mathrm{u}={\mathrm{e}}^{\mathrm{x}}]$$$$={\tan }^{-1}\left(\mathrm{u}\right)+\mathrm{c}={\tan }^{-1}\left({\mathrm{e}}^{\mathrm{x}}\right)+\mathrm{c}$$

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