intergration by substitution

**Geordie-boy** · December 8th 2009, 10:03 PM

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]

**11rdc11** · December 8th 2009, 10:15 PM

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.

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

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

**Prove It** · December 8th 2009, 10:27 PM

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 $\int{\frac{x + x^2}{\sqrt{1 - x^2}}\,dx}$ by substituting $x = \sin{z}$ .

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

So this means that your answer is

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

**Calculus26** · December 8th 2009, 10:27 PM

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