Trig Substitution

Question

I'm trying to find an equivalent integral for $\int {x^5\over \sqrt{1-x^2}}dx$. I substituted $x=sin(\theta)$ for $x$, which gave me $\int {sin^5(\theta)\over cos(\theta)}d(\theta)$; however, this is wrong. Thanks.

Answer 1

With trig substitution, as in U-substitution, you have to update your differential.

For $x = \sin(\theta)$, we have $dx = \cos(\theta) d\theta$.

Substitute the above for $dx$, and use trig integral techniques to finish the integration.

Answer 2

Set $x=sin(\theta)$. Then $dx=cos(\theta) d\theta$

Therefore we get $$\int dx \frac{x^5}{\sqrt{1-x^2}} = \int d\theta \frac {sin^5(\theta)}{cos(\theta)} cos(\theta)$$

Now use $sin^2(\theta) = 1- cos^2(\theta)$ multiple times. We get:

$$\int d\theta {sin^5(\theta)}= \int d\theta ~[sin(\theta) - 2 cos^2(\theta) sin(\theta) + cos^4(\theta) sin(\theta)] \\= -cos(\theta) + \frac{2}{3} cos^3(\theta) - \frac{1}{5} cos^5(\theta)$$

Answer 3

If Trigonometric substitution is not mandatory,

set $\sqrt{1-x^m}=y$

$\implies\dfrac{-mx^{m-1}}{\sqrt{1-x^m}}dx=dy$ and $x^m=1-y^2$

$$\int\dfrac{x^{mn+m-1}}{\sqrt{1-x^m}}dx=-\dfrac1m\int(1-y^2)^ndy$$

$$m=2\implies\int\dfrac{x^{2n+1}}{\sqrt{1-x^2}}dx=-\dfrac12\int(1-y^2)^ndy$$

Here $2n+1=5\iff n=2$