Deepak Scored 45->99%ile with Bounce Back Crack Course. You can do it too!

Evaluate the following integrals:

Question:

Evaluate the following integrals:

$\int \frac{1}{\sqrt{(x-\alpha)(\beta-x)}} d x,(\beta>\alpha)$

Solution:

let $\mathrm{I}=\int \frac{1}{\sqrt{(\mathrm{x}-\mathrm{\alpha})(\beta-\mathrm{x})}} \mathrm{dx},(\operatorname{as} \beta>\alpha)$

$=\int \frac{1}{\sqrt{-x^{2}-x(\alpha+\beta)-\alpha \beta}} d x$

$=\int \frac{1}{\sqrt{-\left[\mathrm{x}^{2}-2 \mathrm{x}\left(\frac{\alpha+\beta}{2}\right)+\left(\frac{\alpha+\beta}{2}\right)^{2}-\left(\frac{\alpha+\beta}{2}\right)^{2}+\alpha \beta\right]}} d \mathrm{x}$

$=\int \frac{1}{\sqrt{-\left[\left(x-\frac{\alpha+\beta}{2}\right)^{2}-\left(\frac{\alpha+\beta}{2}\right)^{2}\right]}} d x$

$=\int \frac{1}{\sqrt{\left[\left(\frac{\beta-\alpha}{2}\right)^{2}-\left(x-\frac{\alpha+\beta}{2}\right)^{2}\right]}} d x \quad[\beta>\alpha]$

Let $(x-(\alpha+\beta) / 2)=t$

$d x=d t$

$I=\int \frac{1}{\sqrt{\left(\frac{\beta-\alpha}{2}\right)^{2}-t^{2}}} d t$

$=\sin ^{-1}\left(\frac{\mathrm{t}}{\frac{\beta-\alpha}{2}}\right)+\mathrm{c}$

$I=\sin ^{-1}\left(2 \frac{x-\frac{\alpha+\beta}{2}}{\beta-\alpha}\right)+c$

$I=\sin ^{-1}\left(\frac{2 x-\alpha-\beta}{\beta-\alpha}\right)$

Leave a comment

None
Free Study Material