int_0^{x} frac{t^2}{sqrt{a^2+t^2}} dt = | vedclass

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

Question

(D) To evaluate $I = \int_0^{x} \frac{t^2}{\sqrt{a^2+t^2}} dt$,we rewrite the numerator as $t^2 = (t^2+a^2) - a^2$. Then,$I = \int_0^{x} \frac{t^2+a^2

Vedclass · Accepted Answer

$\frac{x}{2} \sqrt{a^2+x^2} - \frac{a^2}{2} \log \left| \frac{x+\sqrt{a^2+x^2}}{a} \right|$

Vedclass · Answer

(D) To evaluate $I = \int_0^{x} \frac{t^2}{\sqrt{a^2+t^2}} dt$,we rewrite the numerator as $t^2 = (t^2+a^2) - a^2$. Then,$I = \int_0^{x} \frac{t^2+a^2}{\sqrt{a^2+t^2}} dt - \int_0^{x} \frac{a^2}{\sqrt{a^2+t^2}} dt$. $I = \int_0^{x} \sqrt{a^2+t^2} dt - a^2 \int_0^{x} \frac{1}{\sqrt{a^2+t^2}} dt$. Using the standard integrals $\int \sqrt{a^2+t^2} dt = \frac{t}{2}\sqrt{a^2+t^2} + \frac{a^2}{2} \log|t+\sqrt{a^2+t^2}|$ and $\int \frac{1}{\sqrt{a^2+t^2}} dt = \log|t+\sqrt{a^2+t^2}|$: $I = \left[ \frac{t}{2}\sqrt{a^2+t^2} + \frac{a^2}{2} \log|t+\sqrt{a^2+t^2}| - a^2 \log|t+\sqrt{a^2+t^2}| \right]_0^x$. $I = \left[ \frac{t}{2}\sqrt{a^2+t^2} - \frac{a^2}{2} \log|t+\sqrt{a^2+t^2}| \right]_0^x$. Evaluating at limits: $I = \left( \frac{x}{2}\sqrt{a^2+x^2} - \frac{a^2}{2} \log|x+\sqrt{a^2+x^2}| \right) - \left( 0 - \frac{a^2}{2} \log|a| \right)$. $I = \frac{x}{2}\sqrt{a^2+x^2} - \frac{a^2}{2} \log \left| \frac{x+\sqrt{a^2+x^2}}{a} \right|$.

$\int_0^{x} \frac{t^2}{\sqrt{a^2+t^2}} dt =$

Similar Questions

$\int\limits_0^\infty [2e^{-x}]\, dx$,where $[x]$ denotes the greatest integer function,is equal to:

If $\int_0^3 (3x^2 - 4x + 2) dx = k$, then an integer root of $3x^2 - 4x + 2 = \frac{3k}{5}$ is

The value of $\int_{0}^{\infty} \frac{x}{(1+x)(x^{2}+1)} dx$ is

$\int_{0}^{3} {|2 - x| \, dx}$ equals

If $a_n = \int_0^{\frac{\pi}{2}} \frac{\sin^2 nx}{\sin x} dx$,then $a_2 - a_1, a_3 - a_2, a_4 - a_3$ are in

Vedclass Test Series

Exam Paper Generator

Online Exam Module