Question

integration of root x by root over a minus x dx

Answer 1

To find: $\mathsf{\int \sqrt{\dfrac{x}{a-x}}dx}$

Step-by-step explanation:

Let, $\mathsf{x=a\:sin^{2}\theta}$

$\Rightarrow \mathsf{dx=2a\:sin\theta\:cos\theta\:d\theta}$

Now, $\mathsf{\dfrac{x}{a-x}}$

$\mathsf{=\dfrac{a\:sin^{2}\theta}{a-a\:sin^{2}\theta}}$

$\mathsf{=\dfrac{a\:sin^{2}\theta}{a(1-sin^{2}\theta)}}$

$\mathsf{=\dfrac{sin^{2}\theta}{cos^{2}\theta}}$

• since $\mathsf{sin^{2}\theta+cos^{2}\theta=1}$

$\mathsf{\Rightarrow \dfrac{x}{a-x}=\dfrac{sin^{2}\theta}{cos^{2}\theta}}$

Taking square root to both sides, we get

$\quad \mathsf{\sqrt{\dfrac{x}{a-x}}=\sqrt{\dfrac{sin^{2}\theta}{cos^{2}\theta}}}$

$\Rightarrow \mathsf{\sqrt{\dfrac{x}{a-x}}=\dfrac{sin\theta}{cos\theta}}$

Taking the product of $\mathsf{dx}$ in both sides, we get

$\quad \mathsf{\sqrt{\dfrac{x}{a-x}}dx=\dfrac{sin\theta}{cos\theta}\:dx}$

$\Rightarrow \mathsf{\sqrt{\dfrac{x}{a-x}}dx=\dfrac{sin\theta}{cos\theta}\times 2a\:sin\theta\:cos\theta\:d\theta}$

• since $\mathsf{dx=2a\:sin\theta\:cos\theta\:d\theta}$

$\Rightarrow \mathsf{\sqrt{\dfrac{x}{a-x}}dx=a\times 2\:sin^{2}\theta\:d\theta}$

$\Rightarrow \mathsf{\sqrt{\dfrac{x}{a-x}}dx=a\:(1-cos2\theta)\:d\theta}$

• since $\mathsf{1-2\:sin^{2}\theta=cos2\theta}$

$\Rightarrow \mathsf{\sqrt{\dfrac{x}{a-x}}dx=a\:d\theta-a\:cos2\theta\:d\theta}$

Taking integration to both sides, we get

$\quad \mathsf{\int\sqrt{\dfrac{x}{a-x}}dx=a\:\int d\theta-a\:\int cos2\theta\:d\theta}$

$\Rightarrow \mathsf{\int\sqrt{\dfrac{x}{a-x}}dx=a\:\theta-a\:\dfrac{sin2\theta}{2}+C}$

• where $\mathsf{C}$ is integral constant

$\mathsf{\Rightarrow\int\sqrt{\dfrac{x}{a-x}}dx=a\:sin^{-1}(\sqrt{\dfrac{x}{a}})-\dfrac{a}{2}\times \dfrac{2}{a}\sqrt{x(a-x)}+C}$

• where $\mathsf{\theta=sin^{-1}(\sqrt{\dfrac{x}{a}})}$

• and $\mathsf{sin2\theta=\dfrac{2}{a}\sqrt{x(a-x)}}$

$\mathsf{\Rightarrow\int\sqrt{\dfrac{x}{a-x}}d =a\:sin^{-1}(\sqrt{\dfrac{x}{a}})-\sqrt{x(a-x)}+C}$

This is the required integral.

Answer:

$\mathsf{\int\sqrt{\dfrac{x}{a-x}}d =a\:sin^{-1}(\sqrt{\dfrac{x}{a}})-\sqrt{x(a-x)}+C}$

where $\mathsf{C}$ is integral constant.