Question

If $I=\int xsi{n}^{-1}\left\{\frac{1}{2}\sqrt{\frac{2a-x}{a}}\right\}dx=\frac{A}{248}(x^2-2a^2)co{s}^{-1}\frac{x}{2a}-\frac{1}{8}x\sqrt{4a^2-x^2}+C$ then A is equal to

Answer 1

$I=\int x{\sin }^{-1}\left(\frac{1}{2}\sqrt{\frac{2a-x}{a}}\right)dx$
Substituting $t=\sqrt{2a-x}\Rightarrow dt=-\frac{1}{2\sqrt{2a-x}}dx$
$I=2\int t(t^2-2a){\sin }^{-1}\left(\frac{t}{2\sqrt{t}}\right)dt=2\int t^3{\sin }^{-1}\left(\frac{t}{2\sqrt{t}}\right)dt-4a\int t{\sin }^{-1}\left(\frac{t}{2\sqrt{t}}\right)dt=\frac{1}{2}{t}^4{\sin }^{-1}\left(\frac{t}{2\sqrt{t}}\right)-\frac{1}{4\sqrt{a}}\int \frac{t^4}{\sqrt{1-\frac{t^2}{4a}}}dt-2a{t}^2{\sin }^{-1}\left(\frac{t}{2\sqrt{t}}\right)+\sqrt{a}\int \frac{t^2}{\sqrt{1-\frac{t^2}{4a}}}dt$
Now let $I_1=\frac{1}{4\sqrt{a}}\int \frac{t^4}{\sqrt{1-\frac{t^2}{4a}}}dt$
Substituting $t=2\sqrt{a}\sin u\Rightarrow dt=2\sqrt{a}\cos udu$
$I_1=8a^2\int {\sin }^{4}udu=2a^2{\sin }^{3}u\cos u+\frac{3}{2}{a}^2\sin 2u-6a^{2}u$
And let $I_2=\sqrt{a}\int \frac{t^2}{\sqrt{1-\frac{t^2}{4a}}}dt$
Substituting $t=2\sqrt{a}\sin s\Rightarrow 2\sqrt{a}\cos sds$
$I_2=2a\int 4a{\sin }^{2}sds=-2a^2\sin 2s+4a^{2}s$
Re-substituting values
$I=\frac{1}{8}((4x^2-8a^2){\sin }^{-1}\left(\frac{1}{2}\sqrt{\frac{2a-x}{a}}\right)-\sqrt{a}\sqrt{2a-x}\sqrt{\frac{x}{a}+2})+c=\frac{1}{4}(x^2-2a^2)co{s}^{-1}\frac{x}{2a}-\frac{1}{8}x\sqrt{4a^2-x^2}+c$
$\Rightarrow A=62$