If $b>a$, then $\int_a^b \frac{d x}{\sqrt{(x-a)(b-x)}}$ is equal to

If $b>a$, then $\int_a^b \frac{d x}{\sqrt{(x-a)(b-x)}}$ is equal to
  1. $\pi / 2$
  2. $\pi / 3$
  3. $\pi / 6$
  4. $\pi$

Solution

$\int_a^b \frac{d x}{\sqrt{(x-a)(b-x)}}$ $ \begin{aligned} & \int_a^b \frac{d x}{\sqrt{-x^2+(a+b) x-a b}} \\ & =\int_a^b \frac{d x}{\sqrt{-\left(x^2-(a+b) x+a b\right.}} \\ & =\int_a^b \frac{d x}{\sqrt{-\left[x^2+\left(\frac{a+b}{2}\right)^2+a b-\left(\frac{a+b}{2}\right)^2\right]-(a+b) x}} \end{aligned} $ $\begin{aligned} & =\int_a^b \frac{d x}{\sqrt{\left(\frac{a-b}{2}\right)^2-\left\{x-\left(\frac{a+b}{2}\right)\right\}^2}} \quad[\because b>a] \\ & =\int_a^b \frac{d x}{\sqrt{\left(\frac{b-a}{2}\right)^2-\left\{x-\left(\frac{a+b}{2}\right)\right\}^2}} \\ & \therefore \int \frac{d x}{\sqrt{a^2-x^2}}=\sin ^{-1} \frac{x}{a}+C\end{aligned}$ $\begin{aligned} & =\left\{\sin ^{-1}\left[\frac{x-\left(\frac{a+b}{2}\right)}{\frac{b-a}{2}}\right]\right]_a^b \\ & =\sin ^{-1}\left[\frac{b-\left(\frac{a+b}{2}\right)}{\frac{b-a}{2}}\right]-\sin ^{-1}\left(\frac{a-\frac{a+b}{2}}{\frac{b-a}{2}}\right) \\ & =\sin ^{-1}(+1)-\sin ^{-1}(-1)=+\frac{\pi}{2}+\frac{\pi}{2}=+\pi\end{aligned}$

Asked in: AP EAMCET 2021 (24 Aug Shift 1)

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