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Question Number 112699 by bemath last updated on 09/Sep/20

$$\left(1\right)\int \frac{\sin \mathrm{x}{\mathrm{e}}^{\sqrt{\cos \mathrm{x}}}}{\sqrt{\cos \mathrm{x}}}\mathrm{dx}$$$$\left(2\right)\frac{1+\sqrt{1+\sin 2\mathrm{A}}}{1-\sqrt{1+\sin 2\mathrm{A}}}??$$$$\left(3\right)\int \frac{\mathrm{dx}}{(\mathrm{x}+\mathrm{b})({\mathrm{x}}^2+{\mathrm{a}}^2)}$$

Commented by Dwaipayan Shikari last updated on 09/Sep/20

$$\int \frac{{sinxe}^{\sqrt{cosx}}}{\sqrt{cosx}}dxcosx=t^2,-sinx=2t\frac{dt}{dx}$$$$-\int \frac{2{te}^t}{t}dt=-2e^t+C=-2e^{\sqrt{cosx}}+C$$

Answered by bobhans last updated on 09/Sep/20

$$\left(⧫\right)\int \frac{\sin \mathrm{x}{\mathrm{e}}^{\sqrt{\cos \mathrm{x}}}}{\sqrt{\cos \mathrm{x}}}\mathrm{dx}=\mathrm{I}$$$$\mathrm{let}\mathrm{u}=\sqrt{\cos \mathrm{x}}\to \mathrm{du}=\frac{-\sin \mathrm{x}}{2\sqrt{\cos \mathrm{x}}}\mathrm{dx}$$$$\mathrm{I}=\int {\mathrm{e}}^{\mathrm{u}}(-2\mathrm{du})=-{2\mathrm{e}}^{\mathrm{u}}+\mathrm{c}$$$$\mathrm{I}=-{2\mathrm{e}}^{\sqrt{\cos \mathrm{x}}}+\mathrm{c}$$

Answered by bobhans last updated on 09/Sep/20

$$\left(⧫⧫\right)\frac{1+\sqrt{1+\sin 2\mathrm{A}}}{1-\sqrt{1+\sin 2\mathrm{A}}}=\frac{1+\sin \mathrm{A}+\cos \mathrm{A}}{1-(\sin \mathrm{A}+\cos \mathrm{A})}$$$$=\frac{1+2\sin \left(\frac{\mathrm{A}}{2}\right)\cos \left(\frac{\mathrm{A}}{2}\right)+2\cos {}^2\left(\frac{\mathrm{A}}{2}\right)-1}{1-2\sin \left(\frac{\mathrm{A}}{2}\right)\cos \left(\frac{\mathrm{A}}{2}\right)-(1-2\sin {}^2\left(\frac{\mathrm{A}}{2}\right))}$$$$=\frac{2\cos \left(\frac{\mathrm{A}}{2}\right)(\sin \left(\frac{\mathrm{A}}{2}\right)+\cos \left(\frac{\mathrm{A}}{2}\right))}{2\sin \left(\frac{\mathrm{A}}{2}\right)(\sin \left(\frac{\mathrm{A}}{2}\right)-\cos \left(\frac{\mathrm{A}}{2}\right))}$$$$=\cot \left(\frac{\mathrm{A}}{2}\right).\frac{\tan \left(\frac{\mathrm{A}}{2}\right)+1}{\tan \left(\frac{\mathrm{A}}{2}\right)-1}$$$$$$

Answered by 1549442205PVT last updated on 09/Sep/20

$$\mathrm{P}=\frac{1+\sqrt{1+\sin 2\mathrm{A}}}{1-\sqrt{1+\sin 2\mathrm{A}}}=\frac{1+\sqrt{{(\mathrm{sinA}+\mathrm{cosA})}^2}}{1-\sqrt{(\mathrm{sinA}+\mathrm{cosA})}}$$$$=\frac{1+\mid \mathrm{sinA}+\mathrm{cosA}\mid }{1-\mid \mathrm{sinA}+\mathrm{cosA}\mid }$$$$\mathrm{i})\mathrm{If}\mathrm{sinA}+\mathrm{cosA}⩾0\mathrm{then}$$$$\mathrm{P}=\frac{(1+\mathrm{cosA})+\mathrm{sinA}}{(1+\mathrm{cosA})-\mathrm{sinA}}=\frac{{2\cos }^2\frac{\mathrm{A}}{2}+2\sin \frac{\mathrm{A}}{2}\cos \frac{\mathrm{A}}{2}}{{2\cos }^2\frac{\mathrm{A}}{2}-2\sin \frac{\mathrm{A}}{2}\cos \frac{\mathrm{A}}{2}}$$$$=\frac{2\cos \frac{\mathrm{A}}{2}(\cos \frac{\mathrm{A}}{2}+\sin \frac{\mathrm{A}}{2})}{2\cos \frac{\mathrm{A}}{2}(\cos \frac{\mathrm{A}}{2}-\sin \frac{\mathrm{A}}{2})}=\frac{\cos \frac{\mathrm{A}}{2}+\sin \frac{\mathrm{A}}{2}}{\cos \frac{\mathrm{A}}{2}-\sin \frac{\mathrm{A}}{2}}$$$$=\frac{\sqrt{2}\mathbf{cos}(\frac{\mathit{\pi }}{4}-\frac{\mathbf{A}}{2})}{\sqrt{2}\mathbf{cos}(\frac{\mathit{\pi }}{4}+\frac{\mathbf{A}}{2})}=\frac{\mathbf{cos}(\frac{\mathit{\pi }}{4}-\frac{\mathbf{A}}{2})}{\mathbf{cos}(\frac{\mathit{\pi }}{4}+\frac{\mathbf{A}}{2})}$$$$\mathrm{ii})\mathrm{If}\mathrm{sinA}+\mathrm{cosA}<0\mathrm{then}.\mathrm{Similarly},$$$$\mathrm{we}\mathrm{get}\mathrm{P}=\frac{1-\mathrm{sinA}-\mathrm{cosA}}{1+\mathrm{sinA}+\mathrm{cosA}}$$$$=\frac{{2\sin }^2\frac{\mathrm{A}}{2}-2\sin \frac{\mathrm{A}}{2}\cos \frac{\mathrm{A}}{2}}{{2\cos }^2\frac{\mathrm{A}}{2}+2\sin \frac{\mathrm{A}}{2}\cos \frac{\mathrm{A}}{2}}=\frac{2\sin \frac{\mathrm{A}}{2}(\sin \frac{\mathrm{A}}{2}-\cos \frac{\mathrm{A}}{2})}{2\cos \frac{\mathrm{A}}{2}(\cos \frac{\mathrm{A}}{2}+\sin \frac{\mathrm{A}}{2})}$$$$=\mathbf{tan}\frac{\mathbf{A}}{2}\times \frac{\mathbf{sin}(\frac{\mathit{\pi }}{4}-\frac{\mathbf{A}}{2})}{\mathbf{sin}(\frac{\mathit{\pi }}{4}+\frac{\mathbf{A}}{2})}$$$$3)\mathbf{Find}\int \frac{\mathbf{dx}}{(\mathbf{x}+\mathbf{b})({\mathbf{x}}^2+{\mathbf{a}}^2)}$$$$\mathrm{F}=\int \frac{\mathrm{dx}}{(\mathrm{x}+\mathrm{b})({\mathrm{x}}^2+{\mathrm{a}}^2)}=\int \frac{\mathrm{dx}}{({\mathrm{a}}^2+{\mathrm{b}}^2)(\mathrm{x}+\mathrm{b})}$$$$-\int \frac{\mathrm{x}-\mathrm{b}}{({\mathrm{a}}^2+{\mathrm{b}}^2)({\mathrm{x}}^2+{\mathrm{a}}^2)}\mathrm{dx}$$$$=\frac{1}{{\mathbf{a}}^2+{\mathbf{b}}^2}\mathbf{ln}\mid \mathbf{x}+\mathbf{b}\mid -\frac{1}{{\mathit{a}}^2+{\mathit{b}}^2}\times \frac{1}{2\mathit{a}}{\mathit{t}\mathit{a}\mathit{n}}^{-1}\left(\frac{\mathit{x}}{\mathit{a}}\right)$$

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