Definite integration $\int_0^a\frac{2a-x}{a+x}\sqrt{\frac{a-x}{a+x}}dx$

Question

Find $\int_0^a\frac{2a-x}{a+x}\sqrt{\frac{a-x}{a+x}}dx$. I tried by rationalising the numerator and then replaced $x$ by $a\sin t$, My answer is coming $a(7-2\pi)$ Am I correct? Given ans is a only. That is why I am confused.

Answer 1

Substituting $ \small\left\lbrace\begin{aligned}y&=\sqrt{\frac{a-x}{a+x}}\\ \mathrm{d}x &=-\frac{4ay}{\left(1+y^{2}\right)^{2}}\,\mathrm{d}y\end{aligned}\right. $, we get : \begin{aligned}\int_{0}^{a}{\frac{2a-x}{a+x}\sqrt{\frac{a-x}{a+x}}\,\mathrm{d}x}&=2a\int_{0}^{1}{\frac{y^{2}\left(1+3y^{2}\right)}{\left(1+y^{2}\right)^{2}}\,\mathrm{d}y}\\ &=2a\int_{0}^{1}{\left(3+\frac{2}{\left(1+y^{2}\right)^{2}}-\frac{5}{1+y^{2}}\right)\mathrm{d}y}\\ &=6a+4a\int_{0}^{1}{\frac{\mathrm{d}y}{\left(1+y^{2}\right)^{2}}}-10a\int_{0}^{1}{\frac{\mathrm{d}y}{1+y^{2}}}\\ &=6a+4a\int_{0}^{\frac{\pi}{4}}{\cos^{2}{\theta}\,\mathrm{d}\theta}-a\frac{5\pi}{2}\\ &=6a+4a\left[\frac{\theta+\sin{\theta}\cos{\theta}}{2}\right]_{0}^{\frac{\pi}{4}}-a\frac{5\pi}{2}\\ &=a\left(7-2\pi\right)\end{aligned}

In the fourth line we substituted $ \small\left\lbrace\begin{aligned}y&=\tan{\theta}\\ \mathrm{d}\theta &=\frac{\mathrm{d}y}{1+y^{2}}\end{aligned}\right. \cdot $

Answer 2

Here is to integrate with the substitution $x=a\cos t$,

$$\int_0^a\frac{2a-x}{a+x}\sqrt{\frac{a-x}{a+x}}dx =a\int_0^{\pi/2} (3\tan^2\frac t2 -2\sin^2\frac t2)dt=a( 7-2\pi) $$