Question

Integrate the function ​

Answer 1

Question :

$\displaystyle \bullet\ \; \; \sf \red{ \int \sqrt{\dfrac{1+x}{1-x}}\ dx}$

Solution :

$\displaystyle \sf \int \sqrt{\dfrac{1+x}{1-x}}\ dx$

Rationalizing the denominator ,

$\longrightarrow \displaystyle \sf \int \sqrt{\dfrac{1+x}{1-x} \times \dfrac{1+x}{1+x}}\ dx$

$\longrightarrow \displaystyle \sf \int \sqrt{\dfrac{(1+x)^2}{1-x^2} }\ dx$

$\longrightarrow \displaystyle \sf \int \dfrac{(1+x)}{\sqrt{1-x^2}}\ dx$

$\longrightarrow \displaystyle \sf \int \dfrac{1}{\sqrt{1-x^2}}\ dx+ \int \dfrac{x}{\sqrt{1-x^2}}\ dx$

$\displaystyle \bullet\ \; \; \sf \orange{sin^{-1}\ x= \int \dfrac{1}{\sqrt{1-x^2}}\ dx}$

For second Integral :

Use substitution method ,

➠ u = 1 - x²

➠ du = - 2x dx

➠ $\sf - \dfrac{du}{2}$ = x dx

$\longrightarrow \displaystyle \sf sin^{-1}\ x + \int \dfrac{-du}{2\sqrt{u}}$

$\longrightarrow \displaystyle \sf sin^{-1}\ x -\dfrac{1}{2} \int \dfrac{du}{\sqrt{u}}$

$\longrightarrow \displaystyle \sf sin^{-1}\ x -\dfrac{1}{2} \int u^{-\frac{1}{2}}\ du$

$\longrightarrow \displaystyle \sf sin^{-1}\ x -\dfrac{1}{2} \left[ \dfrac{u^{- \frac{1}{2}+1}}{- \frac{1}{2}+1} \right]+c$

$\longrightarrow \displaystyle \sf sin^{-1}\ x -\dfrac{1}{2} \left[ \dfrac{u^{ \frac{1}{2}}}{\frac{1}{2}} \right]+c$

$\longrightarrow \displaystyle \sf sin^{-1}\ x -u^{\frac{1}{2}}+c$

$\bullet\ \; \; \sf \green{We\ have\ , u=1-x^2}$

$\longrightarrow \displaystyle \sf sin^{-1}\ x -\left( 1-x^2 \right) ^{\frac{1}{2}}+c$

$\longrightarrow \displaystyle \sf \pink{sin^{-1}\ x -\sqrt{(1-x^2)}+c}$