Integrate $1/\sqrt{9-16x^2}$

Question

I need to compute $$ \int \frac{dx}{\sqrt{9-16x^2}} $$ using this table of integrals formula: Table of Integrals

I am getting $\arcsin(x/3) + k$ but I know that is wrong. What do I need to do to correctly solve this problem

Answer 1

Set

$$x = \frac{3}{4}\sin\theta ~~~~~ \text{d}x = \frac{3}{4}\cos\theta\ \text{d}\theta$$

Hence you obtain:

$$\int \frac{\frac{3}{4}\cos\theta\ \text{d}\theta}{\sqrt{9(1 - \sin^2\theta)}} = \int \frac{\frac{3}{4}\cos\theta\ \text{d}\theta}{\sqrt{9\cos^2\theta}} = \int \frac{\frac{3}{4}\cos\theta\ \text{d}\theta}{3\cos\theta} = \frac{1}{4}\int\text{d}\theta = \frac{\theta}{4}$$

Substituting back to $x$, using the substitution we made at the beginning, you have $\theta = \arcsin\left(\frac{4x}{3}\right)$ hence the result is

$$\frac{1}{4}\arcsin\left(\frac{4x}{3}\right)$$

Answer 2

hint: $$x=\frac{3}{4}\sin u$$

Answer 3

HINT

Note that $$ \int \frac{dx}{\sqrt{9-16x^2}} = \int \frac{dx}{\sqrt{3^2-(4x)^2}} $$

Answer 4

$$\sqrt{9-16x^2} = \sqrt{16\left(\frac9{16}-x^2\right)} = 4\sqrt{\frac9{16}-x^2} = 4\sqrt{\left(\frac34\right)^2 - x^2}$$ Now use the table and be careful with handling that factor of $4$ outside of the radical.

Answer 5

Write $$ 9-16x^2=9\left(1-\left(\frac{4x}{3}\right)^2\right) $$ Now set $t=4x/3$, so $dx=\frac{3}{4}\,dt$ and the integral is $$ \frac{3}{4}\int\frac{1}{\sqrt{1-t^2}}\,dt $$ which is elementary.