Solve the integral. ∫v2dv(49−v2)32

Step 1
Given: ${\displaystyle \int \frac{v^{2}dv}{{(49-v^2)}^{\frac{3}{2}}}}$
To find- The value of the above integral.
Concept Used- Above integral can be evaluated by substitution method.
Step 2
Explanation- Rewrite the given expression as,
${\displaystyle I=\int \frac{v^{2}dv}{{(49-v^2)}^{\frac{3}{2}}}}$
substituting ${\displaystyle {(49-v^2)}^{\frac{-1}{2}}=t}$ and differentiating both sides w.r.t. v, we get,
${\displaystyle \frac{-1}{2}\cdot \frac{1}{{(49-v^2)}^{\frac{3}{2}}}\cdot (-2v)dv=dt}$
${\displaystyle \frac{1}{{(49-v^2)}^{\frac{3}{2}}}dv=\frac{dt}{v}}$
From the above expression, we can write as,
${\displaystyle =\int v^2\cdot \frac{dt}{v^2}}$
${\displaystyle =\int dt}$
=t+C
${\displaystyle ={(49-v^2)}^{\frac{-1}{2}}+C}$
${\displaystyle =\frac{1}{\sqrt{(49-v^2)}}+C}$
Answer-Hence the value of the integral ${\displaystyle \int \frac{v^{2}dv}{{(49-v^2)}^{\frac{3}{2}}}is\frac{1}{\sqrt{(49-v^2)}}+C}$.