int frac{dx}{sqrt{x^2 - a^2}} equals | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Let $I = \int \frac{dx}{\sqrt{x^2 - a^2}}$.Substitute $x = a \sec 	heta$,then $dx = a \sec 	heta 	an 	heta \, d	heta$.Substituting these into

Vedclass · Accepted Answer

$\log_e|x + \sqrt{x^2 - a^2}| + c$

Vedclass · Answer

(B) Let $I = \int \frac{dx}{\sqrt{x^2 - a^2}}$.Substitute $x = a \sec 	heta$,then $dx = a \sec 	heta 	an 	heta \, d	heta$.Substituting these into the integral:$I = \int \frac{a \sec 	heta 	an 	heta \, d	heta}{\sqrt{a^2 \sec^2 	heta - a^2}} = \int \frac{a \sec 	heta 	an 	heta \, d	heta}{a 	an 	heta} = \int \sec 	heta \, d	heta$.The integral of $\sec 	heta$ is $\log_e |\sec 	heta + 	an 	heta| + c$.Since $x = a \sec 	heta$,we have $\sec 	heta = \frac{x}{a}$ and $	an 	heta = \sqrt{\sec^2 	heta - 1} = \sqrt{\frac{x^2}{a^2} - 1} = \frac{\sqrt{x^2 - a^2}}{a}$.Therefore,$I = \log_e \left| \frac{x}{a} + \frac{\sqrt{x^2 - a^2}}{a} ight| + c = \log_e \left| \frac{x + \sqrt{x^2 - a^2}}{a} ight| + c$.Using the property $\log(m/n) = \log m - \log n$,we get $\log_e |x + \sqrt{x^2 - a^2}| - \log_e |a| + c$. Since $-\log_e |a| + c$ is just another constant,we write it as $c$.Thus,$I = \log_e |x + \sqrt{x^2 - a^2}| + c$.

$\int \frac{dx}{\sqrt{x^2 - a^2}}$ equals

Similar Questions

If $\int \frac{dx}{1 + \sin x} = \tan \left( \frac{x}{2} + a \right) + b$,then

$\int {\left( {1 + x + \frac{{{x^2}}}{{2!}} + \frac{{{x^3}}}{{3!}} + \dots} \right) dx} = $

$\int \frac{\sin \frac{5x}{2}}{\sin \frac{x}{2}} dx = $ (where $C$ is a constant of integration.)

$\int {(\sin^4 x - \cos^4 x) \, dx} = $

$\int \frac{\cos x+x \sin x}{x(x+\cos x)} d x=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module