How do you integrate $\int \frac{3 x}{\sqrt{(1 - x^{2})}} d x$ $int (3x)/sqrt((1-x^2))dx$ using trigonometric substitution?

Question

How do you find the integral $\int \frac{1}{x^{2} \cdot \sqrt{x^{2} - 9}} d x$ $int1/(x^2*sqrt(x^2-9))dx$ ?
How do you find the integral $\int \frac{x^{3}}{\sqrt{x^{2} + 9}} d x$ $intx^3/(sqrt(x^2+9))dx$ ?
How do you find the integral $\int x^{3} \cdot \sqrt{9 - x^{2}} d x$ $intx^3*sqrt(9-x^2)dx$ ?
How do you find the integral $\int \frac{x^{3}}{\sqrt{16 - x^{2}}} d x$ $intx^3/(sqrt(16-x^2))dx$ ?
How do you find the integral $\int \frac{\sqrt{x^{2} - 1}}{x} d x$ $intsqrt(x^2-1)/xdx$ ?
How do you find the integral $\int \frac{\sqrt{x^{2} - 9}}{x^{3}} d x$ $intsqrt(x^2-9)/x^3dx$ ?
How do you find the integral $\int \frac{x}{\sqrt{x^{2} + x + 1}} d x$ $intx/(sqrt(x^2+x+1))dx$ ?
How do you find the integral $\int \frac{d t}{\sqrt{t^{2} - 6 t + 13}}$ $intdt/(sqrt(t^2-6t+13))$ ?
How do you find the integral $\int x \cdot \sqrt{1 - x^{4}} d x$ $intx*sqrt(1-x^4)dx$ ?
How do you prove the integral formula $\int \frac{d x}{\sqrt{x^{2} + a^{2}}} = ln (x + \sqrt{x^{2} + a^{2}}) + C$ $intdx/(sqrt(x^2+a^2)) = ln(x+sqrt(x^2+a^2))+ C$ ?

6119 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-01-24T06:28:39

You need to use the beautiful Pythagorean right triangle.

Here is the general antiderivative :
$- 3 \cdot \sqrt{1 - x^{2}}$ $-3*sqrt(1- x^2)$

Here's a quick paint job to show you the lovely right triangle:

enter image source here

You now need to find the following:

$1) x$ $1)x$

$2) \frac{d x}{d θ}$ $2)dx/(d theta)$

$3) \sqrt{1 - x^{2}}$ $3)sqrt(1-x^2)$

I'll give you a gentle and loving push in the right direction:

What is x?

Well, it sure looks to be :

$x = sin (θ)$ $x = sin (theta)$

Your turn!

How do you integrate ∫3x√(1−x2)dxint (3x)/sqrt((1-x^2))dx using trigonometric substitution?