int frac{d x}{left(x^2-a^2right)^{frac{3}{2}} | vedclass

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

Question

(C) Let $I = \int \frac{dx}{(x^2 - a^2)^{3/2}}$.Substitute $x = a \sec 	heta$, then $dx = a \sec 	heta 	an 	heta d	heta$.Substituting these into

Vedclass · Accepted Answer

$-\frac{x}{a^2 \sqrt{x^2-a^2}}+C$

Vedclass · Answer

(C) Let $I = \int \frac{dx}{(x^2 - a^2)^{3/2}}$.Substitute $x = a \sec 	heta$, then $dx = a \sec 	heta 	an 	heta d	heta$.Substituting these into the integral:$I = \int \frac{a \sec 	heta 	an 	heta d	heta}{(a^2 	an^2 	heta)^{3/2}} = \int \frac{a \sec 	heta 	an 	heta d	heta}{a^3 	an^3 	heta} = \frac{1}{a^2} \int \frac{\sec 	heta}{	an^2 	heta} d	heta$.Using $\sec 	heta = \frac{1}{\cos 	heta}$ and $	an 	heta = \frac{\sin 	heta}{\cos 	heta}$, we get:$I = \frac{1}{a^2} \int \frac{1}{\cos 	heta} \cdot \frac{\cos^2 	heta}{\sin^2 	heta} d	heta = \frac{1}{a^2} \int \frac{\cos 	heta}{\sin^2 	heta} d	heta$.Let $u = \sin 	heta$, then $du = \cos 	heta d	heta$. So,$I = \frac{1}{a^2} \int u^{-2} du = \frac{1}{a^2} (-u^{-1}) + C = -\frac{1}{a^2 \sin 	heta} + C$.Since $x = a \sec 	heta$, $\sec 	heta = \frac{x}{a}$, so $\cos 	heta = \frac{a}{x}$.Then $\sin 	heta = \sqrt{1 - \cos^2 	heta} = \sqrt{1 - \frac{a^2}{x^2}} = \frac{\sqrt{x^2 - a^2}}{x}$.Substituting $\sin 	heta$ back, we get:$I = -\frac{1}{a^2 \left( \frac{\sqrt{x^2 - a^2}}{x} ight)} + C = -\frac{x}{a^2 \sqrt{x^2 - a^2}} + C$.

$\int \frac{d x}{\left(x^2-a^2\right)^{\frac{3}{2}}}$ is equal to

Similar Questions

Let $I_n = \int \sec^n x \, dx$. If $5 I_6 - 4 I_4 = f(x)$, then $f\left(\frac{\pi}{4}\right)$ is equal to

If $n$ is a positive integer greater than $1$ and $I_{n}=\int \frac{\sin n x}{\sin x} d x$, then $I_{n+1}-I_{n-1}=$

If $\int \frac{2 x^{12}+5 x^9}{\left(1+x^3+x^5\right)^3} d x=\frac{x^m}{l\left(1+x^3+x^5\right)^r}+C$ then $\frac{m-l}{r}=$

$ \int \frac{1}{x^{2}\left(x^{4}+1\right)^{3 / 4}} dx $ is equal to

If $\int \frac{dx}{(x-1)^{3/2}(x-3)^{1/2}} = \sqrt{f(x)} + c$,then $f(-1) - f(0) =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module