Evaluation of $ \int_{0}^{2a}x\cdot \sin^{-1}\left(\frac{1}{2}\sqrt{\frac{2a-x}{a}}\right)dx$ 
$(1)$ Evaluation of $\displaystyle \int_{0}^{\frac{\pi}{2}}\frac{a\sin x+b\cos x}{\sin \left(x+\frac{\pi}{4}\right)}dx$
$(2)$ Evaluation of $\displaystyle \int_{-1}^{1}\ln\left(\frac{1+x}{1-x}\right)\cdot \frac{x^3}{\sqrt{1-x^2}}dx$
$(3)$ Evaluation of $\displaystyle \int_{0}^{2a}x\cdot \sin^{-1}\left(\frac{1}{2}\sqrt{\frac{2a-x}{a}}\right)dx$

$\bf{My\; Try::}$ For $(1)$ one
Let $\displaystyle I = \int_{0}^{\frac{\pi}{2}}\frac{a\sin x}{\sin (x+\frac{\pi}{4})}dx+\int_{0}^{\frac{\pi}{2}}\frac{b\cos x}{\sin (x+\frac{\pi}{4})}dx$
Now Let $\displaystyle J = \int_{0}^{\frac{\pi}{2}}\frac{a\sin x}{\sin (x+\frac{\pi}{4})}dx$ and $\displaystyle K = \int_{0}^{\frac{\pi}{2}}\frac{b\cos x}{\sin (x+\frac{\pi}{4})}dx$
Now We will Calculate $\displaystyle J = \int_{0}^{\frac{\pi}{2}}\frac{a\sin x}{\sin (x+\frac{\pi}{4})}dx$
Using $\displaystyle \left(x+\frac{\pi}{4}\right)=t\;,$ Then $dx = dt$.
So $\displaystyle J = a\int_{0}^{\frac{\pi}{4}}\frac{\sin \left(t-\frac{\pi}{4}\right)}{\sin t}dt = a\cdot \frac{1}{\sqrt{2}}\cdot \frac{\pi}{2}$
Similarly we will Calculate $\displaystyle K = \int_{0}^{\frac{\pi}{2}}\frac{b\cos x}{\sin (x+\frac{\pi}{4})}dx$
Using $\displaystyle \left(x+\frac{\pi}{4}\right)=t\;,$ Then $dx = dt$.
So $\displaystyle K = b\int_{0}^{\frac{\pi}{4}}\frac{\cos \left(t-\frac{\pi}{4}\right)}{\sin t}dt = b\cdot \frac{1}{\sqrt{2}}\cdot \frac{\pi}{2}$
So $\displaystyle I = J+K = \frac{\pi}{2\sqrt{2}}\cdot (a+b)$
Is there is any Shorter Solution for $(1)$ one and How can I calcultae $(2)$ and $(3)$ one
Help me
Thanks
 A: (1)
We have:
$$\begin{eqnarray*}\int_{0}^{\pi/4}\frac{\sin x}{\sin(x+\pi/4)}\,dx &=& \int_{\pi/4}^{\pi/2}\frac{\sin(x-\pi/4)}{\sin x}\,dx = \frac{1}{\sqrt{2}}\int_{\pi/4}^{\pi/2}\left(1-\cot x\right)\,dx\\&=&\frac{1}{4\sqrt{2}}\left(\pi-\log 4\right),\end{eqnarray*}$$
$$\begin{eqnarray*}\int_{0}^{\pi/4}\frac{\cos x}{\sin(x+\pi/4)}\,dx &=& \int_{\pi/4}^{\pi/2}\frac{\cos(x-\pi/4)}{\sin x}\,dx = \frac{1}{\sqrt{2}}\int_{\pi/4}^{\pi/2}\left(\cot x+1\right)\,dx\\&=&\frac{1}{4\sqrt{2}}\left(\pi+\log 4\right).\end{eqnarray*}$$
(2) We have:
$$\begin{eqnarray*}\int_{-1}^{1}\log\left(\frac{1+x}{1-x}\right)\frac{x^3}{\sqrt{1-x^2}}=2\int_{0}^{\pi/2}\cos^3\theta\cdot\log\left(\frac{1+\cos\theta}{1-\cos\theta}\right)d\theta\end{eqnarray*}$$
and integrating by parts we get:
$$\begin{eqnarray*}\int_{-1}^{1}\log\left(\frac{1+x}{1-x}\right)\frac{x^3}{\sqrt{1-x^2}}=\int_{0}^{\pi/2}\left(3+\frac{\sin(3\theta)}{3\sin\theta}\right)d\theta=\frac{5\pi}{3}.\end{eqnarray*}$$
(3) We have:
$$\begin{eqnarray*}\int_{0}^{2a}x\cdot\arcsin\left(\frac{1}{2}\sqrt{\frac{2a-x}{a}}\right)dx&=&a^2\cdot\int_{0}^{2}x\cdot\arcsin\left(\frac{\sqrt{2-x}}{2}\right)dx\\&=&4a^2\cdot\int_{0}^{1}x\cdot\arcsin\sqrt{\frac{1-x}{2}}\,dx\\&=&2a^2\int_{0}^{\pi/2}\theta \sin\theta\cos\theta\,d\theta\\&=&\frac{\pi a^2}{4}.\end{eqnarray*}$$
A: We will work on the third one.
Set $x=2a t$ then the third one becomes 
$$I_3=\int_{0}^{2a}x\cdot \sin^{-1}\left(\frac{1}{2}\sqrt{\frac{2a-x}{a}}\right)dx=(2a)^2\int_{0}^{1}\sin^{-1}\left(\sqrt{\frac{1-t}{2}}\right)tdt \tag{1}$$
Set $\sin s=\sqrt{(1-t)/2}$, then $t=\cos(2s)$, $dt=-2\sin(2s)ds$. So (1) becomes
$$I_3=(2a)^2\int_{0}^{\pi/4}\sin^{-1}\left(\sin s\right)\cos(2s)2\sin(2s)ds \implies $$
$$ I_3=(2a)^2\int_{0}^{\pi/4}s\sin(4s)ds\tag{2}$$
Set $u=4s$ in (2) we obtain:
$$ I_3=(a/2)^2\int_{0}^{\pi}u\sin udu=(a/2)^2 \pi \tag{3}$$
