Question

plz someone solve it
it's urgent!!!​

Answer 1

Answer:

Step-by-step explanation:

Answer 2

We're asked to evaluate,

$\displaystyle\longrightarrow I=\int\limits_0^a\sin^{-1}\left(\sqrt{\dfrac{x}{a+x}}\right)\ dx$

We can change the sine inverse function to tan inverse as,

• $\sin^{-1}\left(\dfrac{a}{b}\right)=\tan^{-1}\left(\dfrac{a}{\sqrt{b^2-a^2}}\right)$

Taking $a=\sqrt x$ and $b=\sqrt{a+x},$

• $\sin^{-1}\left(\dfrac{\sqrt x}{\sqrt{x+a}}\right)=\tan^{-1}\left(\dfrac{\sqrt x}{\sqrt{a+x-x}}\right)$

• $\sin^{-1}\left(\sqrt{\dfrac{x}{x+a}}\right)=\tan^{-1}\left(\sqrt{\dfrac{x}{a}}\right)$

Then our integral becomes,

$\displaystyle\longrightarrow I=\int\limits_0^a\tan^{-1}\left(\sqrt{\dfrac{x}{a}\right)\ dx$

Let,

$\longrightarrow u=\sqrt{\dfrac{x}{a}}$

$\longrightarrow du=\dfrac{1}{2\sqrt{\dfrac{x}{a}}}\cdot\dfrac{1}{a}\ dx$

$\longrightarrow du=\dfrac{1}{2a}\sqrt{\dfrac{a}{x}}\ dx$

$\longrightarrow dx=2a\sqrt{\dfrac{x}{a}}\ du$

$\longrightarrow dx=2au\ du$

For $x=0,$

$\longrightarrow u=\sqrt{\dfrac{0}{a}}$

$\longrightarrow u=0$

For $x=a,$

$\longrightarrow u=\sqrt{\dfrac{a}{a}}$

$\longrightarrow u=1$

Then our integral becomes,

$\displaystyle\longrightarrow I=\int\limits_0^1\tan^{-1}u\cdot 2au\ du$

$\displaystyle\longrightarrow I=2a\int\limits_0^1\tan^{-1}u\cdot u\ du$

Integrating by parts,

$\displaystyle\longrightarrow I=2a\left[\tan^{-1}u\cdot \int u\ du-\int\dfrac{d}{du}\left(\tan^{-1}u\right)\cdot\left(\int u\ du\right)\ du\right]_0^1$

$\displaystyle\longrightarrow I=2a\left[\tan^{-1}u\cdot\dfrac{u^2}{2}-\int\dfrac{1}{1+u^2}\cdot\dfrac{u^2}{2}\ du\right]_0^1$

$\displaystyle\longrightarrow I=a\left[u^2\tan^{-1}u-\int\dfrac{u^2}{1+u^2}\ du\right]_0^1$

$\displaystyle\longrightarrow I=a\left[u^2\tan^{-1}u-\int\dfrac{1+u^2-1}{1+u^2}\ du\right]_0^1$

$\displaystyle\longrightarrow I=a\left[u^2\tan^{-1}u-\int\left(1-\dfrac{1}{1+u^2}\right)\ du\right]_0^1$

$\displaystyle\longrightarrow I=a\left[u^2\tan^{-1}u-u+\tan^{-1}u\right]_0^1$

$\displaystyle\longrightarrow I=a\left(\tan^{-1}(1)-1+\tan^{-1}1\right)$

$\displaystyle\longrightarrow\underline{\underline{I=a\left(\dfrac{\pi}{2}-1\right)}}$