Question

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Answer 1

$\displaystyle\underline{\text{Solution :}}$

$\displaystyle \mathrm{Now,\:\int \frac{sin^{2}\theta\:d\theta}{m+H\:sin\theta\:cos\theta}}$

$\displaystyle \mathrm{=\frac{2}{2}\int \frac{1-(1-2\:sin^{2}\theta)}{2m+2H\:sin\theta\:cos\theta}d\theta}$

$\displaystyle \mathrm{=\int \frac{1-cos2\theta}{2m+2H\:sin\theta\:cos\theta}d\theta}$

$\displaystyle \mathrm{=\frac{1}{2}\int \frac{d\theta}{m+H\:sin\theta\:cos\theta}-\int \frac{cos2\theta\:d\theta}{2m+H\:sin2\theta}}$

$\displaystyle \mathrm{=\frac{1}{2}\int \frac{sec^{2}\theta\:d\theta}{m\:sec^{2}\theta+H\:tan\theta}-\frac{1}{2}\int \frac{2cos2\theta\:d\theta}{2m+H\:sin2\theta}.....(i)}$

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$\displaystyle \mathrm{Let,\:tan\theta=y}$

$\displaystyle\to \mathrm{sec^{2}\theta\:d\theta=dy}$

$\displaystyle \mathrm{Also,\:sec^{2}\theta=1+tan^{2}\theta=1+y^{2}}$

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$\displaystyle \mathrm{Let,\:sin2\theta=z}$

$\displaystyle\to \mathrm{2cos2\theta\:d\theta=dz}$

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$\displaystyle\text{Continuing (i)}\implies$

$\displaystyle\mathrm{=\frac{1}{2}\int \frac{dy}{m(y^{2}+1)+Hy}-\frac{1}{2}\int \frac{dz}{2m+Hz}}$

$\displaystyle\mathrm{=\frac{1}{2m}\int \frac{dy}{y^{2}+1+\frac{H}{m}y}-\frac{1}{2H}\int \frac{dz}{z+\frac{2m}{H}}}$

$\displaystyle\mathrm{=\frac{1}{2m}\int \frac{dy}{(y+\frac{H}{2m})^{2}+(\sqrt{1-\frac{H^{2}}{4m^{2}}})^{2}}-\frac{1}{2H}\int \frac{d(z+\frac{2m}{H})}{z+\frac{2m}{H}}}$

$\displaystyle\mathrm{=\frac{1}{2m}\frac{1}{\sqrt{1-\frac{H^{2}}{4m^{2}}}}tan^{-1}\frac{y+\frac{H}{2m}}{\sqrt{1-\frac{H^{2}}{4m^{2}}}}-\frac{1}{2H}log(z+\frac{2m}{H})}$

$\displaystyle\mathrm{=\frac{1}{2m}\frac{1}{\frac{\sqrt{4m^{2}-H^{2}}}{2m}}tan^{-1}\frac{\frac{2my+H}{2m}}{\frac{\sqrt{4m^{2}-H^{2}}}{2m}}-\frac{1}{2H}log(\frac{Hz+2m}{H})}$

$\displaystyle\mathrm{=\frac{1}{\sqrt{4m^{2}-H^{2}}}tan^{-1}(\frac{2m\:tan\theta}{\sqrt{4m^{2}-H^{2}}})-\frac{1}{2H}log(\frac{H\:sin2\theta}{H})}$

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$\displaystyle\mathrm{Then,\:a=Mg\int_{\frac{\pi}{4}}^{0}\frac{sin^{2}\theta}{m+H\:sin\theta\:cos\theta}}$

$\displaystyle\mathrm{=\frac{Mg}{\sqrt{4m^{2}-H^{2}}}\bigg[tan^{-1}\frac{2m\:tan\theta+H}{\sqrt{4m^{2}-H^{2}}}\bigg]_{\frac{\pi}{4}}^{0}-\frac{Mg}{2H}\bigg[log\frac{H\:sin2\theta}{H}\bigg]_{\frac{\pi}{4}}^{0}}$

$\displaystyle\mathrm{=\frac{Mg}{\sqrt{4m^{2}-H^{2}}}\bigg[tan^{-1}\frac{H}{\sqrt{4m^{2}-H^{2}}}-tan^{-1}\frac{2m+H}{\sqrt{4m^{2}-H^{2}}}\bigg]-\frac{Mg}{2H}\bigg[log\frac{2m}{H}-log\frac{H+2m}{H}\bigg]}$

$\displaystyle\mathrm{=\frac{Mg}{\sqrt{4m^{2}-H^{2}}}tan^{-1}\bigg[\frac{\frac{H-2m-H}{\sqrt{4m^{2}-H^{2}}}}{1+\frac{H(2m+H)}{4m^{2}-H^{2}}}\bigg]-\frac{Mg}{2H}log\bigg[\frac{\frac{2m}{H}}{\frac{H+2m}{H}}\bigg]}$

$\displaystyle\mathrm{=\frac{Mg}{\sqrt{4m^{2}-H^{2}}}tan^{-1}\frac{-2m\sqrt{4m^{2}-H^{2}}}{4m^{2}-H^{2}+2mH+H^{2}}-\frac{Mg}{2H}log\frac{2m}{H+2m}}$

$\displaystyle\mathrm{=\frac{Mg}{\sqrt{4m^{2}-H^{2}}}tan^{-1}\frac{-\sqrt{4m^{2}-H^{2}}}{2m+H}-\frac{Mg}{2H}log\frac{2m}{2m+H}}$

$\displaystyle\mathrm{=-\frac{Mg}{\sqrt{4m^{2}-H^{2}}}tan^{-1}\frac{\sqrt{4m^{2}-H^{2}}}{2m+H}-\frac{Mg}{2H}log\frac{2m}{2m+H}}$

$\displaystyle\mathrm{=-\frac{Mg}{\sqrt{4m^{2}-H^{2}}}tan^{-1}\sqrt{\frac{2m-H}{2m+H}}-\frac{Mg}{2H}log\frac{2m}{2m+H}}$

$\displaystyle\implies \boxed{\mathrm{a=Mg\int_{\frac{\pi}{4}}^{0}\frac{sin^{2}\theta}{m+H\:sin\theta\:cos\theta}=-\frac{Mg}{\sqrt{4m^{2}-H^{2}}}tan^{-1}\sqrt{\frac{2m-H}{2m+H}}-\frac{Mg}{2H}log\frac{2m}{2m+H}}}$