1. ## intergration by substitution

Hi all im trying to get my head around intergration by substitution and this one problem in particular has been giving me a problems any help?

substitute x = sine z into [int(x+x^2)/(1-x^2)^1/2 dx]

2. Originally Posted by Geordie-boy
Hi all im trying to get my head around intergration by substitution and this one problem in particular has been giving me a problems any help?

substitute x = sine z into [int(x+x^2)/(1-x^2)^1/2 dx]
separate the integral into two integrals.

$\displaystyle \int \frac{x}{\sqrt{1-x^2}} + \int\frac{x^2}{\sqrt{1-x^2}}$

for the first integral just use the sub of $\displaystyle u= 1 - x^2$ for the second use trig sub

3. Originally Posted by Geordie-boy
Hi all im trying to get my head around intergration by substitution and this one problem in particular has been giving me a problems any help?

substitute x = sine z into [int(x+x^2)/(1-x^2)^1/2 dx]
Easy.

You're trying to solve $\displaystyle \int{\frac{x + x^2}{\sqrt{1 - x^2}}\,dx}$ by substituting $\displaystyle x = \sin{z}$.

Notice that if you try to do this, you need to change the integral so it is with respect to $\displaystyle z$ instead of $\displaystyle x$.

To do this, if $\displaystyle x = \sin{z}$ then $\displaystyle \frac{dx}{dz} = \cos{z}$.

Thus $\displaystyle dx = \cos{z}\,dz$.

Now substitute this all into the original integral, so that it becomes

$\displaystyle \int{\frac{\sin{z} + \sin^2{z}}{\sqrt{1 - \sin^2{z}}}\,\cos{z}\,dz}$

$\displaystyle = \int{\frac{\sin{z} + \sin^2{z}}{\sqrt{\cos^2{z}}}\,\cos{z}\,dz}$

$\displaystyle = \int{\frac{\sin{z} + \sin^2{z}}{\cos{z}}\,\cos{z}\,dz}$

$\displaystyle = \int{\sin{z} + \sin^2{z}\,dz}$

$\displaystyle = \int{\sin{z} + \frac{1}{2} - \frac{1}{2}\cos{(2z)}\,dz}$

$\displaystyle = -\cos{z} + \frac{1}{2}z - \frac{1}{4}\sin{(2z)} + C$

$\displaystyle = \frac{1}{2}z - \frac{1}{4}\sin{(2z)} - \cos{z} + C$.

Now use the fact that $\displaystyle x = \sin{z}$ to get the answer back into terms of $\displaystyle x$.

If $\displaystyle x = \sin{z}$ then $\displaystyle z = \arcsin{x}$.

Also $\displaystyle x^2 = \sin^2{z}$

$\displaystyle x^2 = 1 - \cos^2{z}$

$\displaystyle x^2 - 1 = -\cos^2{z}$

$\displaystyle 1 - x^2 = \cos^2{z}$

$\displaystyle \sqrt{1 - x^2} = \cos{z}$.

Finally, $\displaystyle \sin{(2z)} = 2\sin{z}\cos{z}$

$\displaystyle = 2x\sqrt{1 - x^2}$.

$\displaystyle \frac{1}{2}\arcsin{z} - \frac{1}{2}x\sqrt{1 - x^2} - \sqrt{1 - x^2} + C$.