# Thread: Integratimg using trig substitution

1. ## Integratimg using trig substitution

need some help doing this question..

$\int\frac{x}{\sqrt(4-x^2)}dx$ given the substitution x=2sint

using
$4 (1 - \sin^2 x) = 4 cos^2x$

I've managed to rewrite the main integral as

$\int\frac{2sint}{\sqrt(4cos^2t)}dx$
then get stuk

2. Originally Posted by Kevlar
need some help doing this question..

$\int\frac{x}{\sqrt(4-x^2)}dx$ given the substitution x=2sint

using
$4 (1 - \sin^2 x) = 4 cos^2x$

I've managed to rewrite the main integral as

$\int\frac{2sint}{\sqrt(4cos^2t)}dx$
then get stuk
Don't forget, if $x = 2\sin{t}$

then $\frac{dx}{dt} = 2\cos{t}$.

So $dx = 2\cos{t}\,dt$.

$\int{\frac{x}{\sqrt{4 - x^2}}\,dx} = \int{\frac{2\sin{t}}{\sqrt{4(1 - \sin^2{t})}}\,2\cos{t}\,dt}$

$= \int{\frac{4\sin{t}\cos{t}}{2\sqrt{\cos^2{t}}}\,dt }$

$= \int{\frac{2\sin{t}\cos{t}}{\cos{t}}\,dt}$

$= \int{2\sin{t}\,dt}$

$= -2\cos{t} + C$.

And since $x = 2\sin{t}$

$x^2 = 4\sin^2{t}$

$\frac{x^2}{4} = \sin^2{t}$

$\frac{x^2}{4} = 1 - \cos^2{t}$

$\cos^2{t} = 1 - \frac{x^2}{4}$

$\cos^2{t} = \frac{4 - x^2}{4}$

$\cos{t} = \frac{\sqrt{4 - x^2}}{2}$.

$-2\cos{t} + C = \frac{-2\sqrt{4 - x^2}}{2} + C$

$=C -\sqrt{4 - x^2}$.

3. Originally Posted by Kevlar
need some help doing this question..

$\int\frac{x}{\sqrt(4-x^2)}dx$ given the substitution x=2sint

using
$4 (1 - \sin^2 x) = 4 cos^2x$

I've managed to rewrite the main integral as

$\int\frac{2sint}{\sqrt(4cos^2t)}dx$
then get stuk
You have write $dx$ in terms of $t$.

$dx=2cos(t)dt$

$\int\frac{2sint}{\sqrt{4cos^2t}}=\int\frac{4sin(t) cos(t)}{2cos(t)}dt=2\int sin(t)dt$