Question

Evaluate: $\int \frac{1}{ \sqrt{9-25x^{2}}}\ dx$

Answer 1

I think it's correct

Answer 2

Answer:

$\dfrac 15 sin^{-1} ({\dfrac {5x}3})$

Step-by-step explanation:

We have to find the integration of $\dfrac 1{(9 - 25x^2)}$

$\int \dfrac {1}{\sqrt {9-25x^2}}dx$ ----equation 1

Substitute

$u = \dfrac {5x}3$

$\dfrac {du}{dx} = \frac 53\\dx = \dfrac 35 du$

Now substitute the values in equation 1

$\int \dfrac {3}{5\sqrt {9 - 9 u^2}}du\\\dfrac 35\int \dfrac{du}{3\sqrt{(1-u^2)}}\\\\ \dfrac 15 \int \dfrac {du}{\sqrt{1 - u^2}}$

This is the definite integral of $sin^{-1}u$

$\dfrac 15 sin^{-1}u + C$

Now, put the value of u

$\dfrac 15 sin^{-1} ({\dfrac {5x}3})$