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Question Number 123154 by liberty last updated on 23/Nov/20

Answered by benjo_mathlover last updated on 23/Nov/20

$$lett=\sin q\Rightarrow dt=\cos qdq$$$$\eta \left(x\right)=\int \sqrt{1-\sin {}^{2}q}\left(\cos qdq\right)$$$$\eta \left(x\right)=\int \left(\frac{1+\cos 2q}{2}\right)dq$$$$\eta \left(x\right)=\frac{q+\frac{1}{2}\sin 2q}{2}+c$$$$=\frac{{\sin }^{-1}\left(t\right)+t\sqrt{1-t^2}}{2}+c$$$$thus\eta =\underset{1}{\stackrel{\cosh \left(x\right)}{\int }}\sqrt{1-t^2}dt$$$$={\left[\frac{{\sin }^{-1}\left(t\right)+t\sqrt{1-t^2}}{2}\right]}_1^{\cosh \left(x\right)}$$$$=\left[\frac{{\sin }^{-1}\left(\cosh \left(x\right)\right)+\cosh \left(x\right)\sinh \left(x\right)}{2}\right]-\frac{\pi }{4}$$$$$$

Answered by Dwaipayan Shikari last updated on 23/Nov/20

$$\int cos\theta \sqrt{1-{sin}^2\theta }d\theta$$$$=\int {cos}^2\theta d\theta =\int \frac{1}{2}+\frac{cos2\theta }{2}d\theta =\frac{\theta }{2}+\frac{sin2\theta }{4}=\frac{{sin}^{-1}t}{2}+\frac{t\sqrt{1-t^2}}{2}$$$${\int }_1^{coshx}\sqrt{1-t^2}dt=\frac{{sin}^{-1}\left(\frac{e^x-e^{-x}}{2}\right)}{2}+\frac{coshx\sqrt{1-{cosh}^{2}x}}{2}-\frac{\pi }{4}$$

Answered by mathmax by abdo last updated on 23/Nov/20

$$\mathrm{f}\left(\mathrm{x}\right)={\int }_1^{\frac{{\mathrm{e}}^{\mathrm{x}}+{\mathrm{e}}^{-\mathrm{x}}}{2}}\sqrt{1-{\mathrm{t}}^2}\mathrm{dt}\Rightarrow \mathrm{f}\left(\mathrm{x}\right){=}_{\mathrm{t}=\sin \theta }{\int }_{\frac{\pi }{2}}^{\arcsin \left(\frac{{\mathrm{e}}^{\mathrm{x}}+{\mathrm{e}}^{-\mathrm{x}}}{2}\right)}\cos \theta \cos \theta \mathrm{d}\theta$$$$=\frac{1}{2}{\int }_{\frac{\pi }{2}}^{\arcsin \left(\frac{{\mathrm{e}}^{\mathrm{x}}+{\mathrm{e}}^{-\mathrm{x}}}{2}\right)}(1+\cos \left(2\theta \right))\mathrm{d}\theta$$$$=\frac{1}{2}{\left[\theta \right]}_{\frac{\pi }{2}}^{\arcsin \left(\frac{{\mathrm{e}}^{\mathrm{x}}+{\mathrm{e}}^{-\mathrm{x}}}{2}\right)}+\frac{1}{4}{\left[\sin \left(2\theta \right)\right]}_{\frac{\pi }{2}}^{\arcsin \left(\frac{{\mathrm{e}}^{\mathrm{x}}+{\mathrm{e}}^{-\mathrm{x}}}{2}\right)}$$$$=\frac{1}{2}\{\arcsin \left(\frac{{\mathrm{e}}^{\mathrm{x}}+{\mathrm{e}}^{-\mathrm{x}}}{2}\right)-\frac{\pi }{2})+\frac{1}{2}{\left[\sin \theta \sqrt{1-{\sin }^2\theta }\right]}_{\frac{\pi }{2}}^{\arcsin \left(\frac{{\mathrm{e}}^{\mathrm{x}}+{\mathrm{e}}^{-\mathrm{x}}}{2}\right)}$$$$\mathrm{f}\left(\mathrm{x}\right)=\frac{1}{2}\left(\arcsin \left(\frac{{\mathrm{e}}^{\mathrm{x}}+{\mathrm{e}}^{-\mathrm{x}}}{2}\right)\right)-\frac{\pi }{4}+\frac{1}{2}\left(\frac{{\mathrm{e}}^{\mathrm{x}}+{\mathrm{e}}^{-\mathrm{x}}}{2}\sqrt{1-{\left(\frac{{\mathrm{e}}^{\mathrm{x}}+{\mathrm{e}}^{-\mathrm{x}}}{2}\right)}^2}\right)$$$$$$

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