Evaluate $\int(x\sqrt{1-x^4})dx$ I've attempted this question with the substitutions $x=\sin(\theta)$ and $u = \sin^2(\theta)$ but then I got stuck. I think the main problem here is the power is too high. I'm not sure how to reduce the power.  Here's the question:   

Evaluate the integral using inverse/trigonometric substitutions:
  $$\int(x\sqrt{1-x^4})dx.$$

 A: Hint: $~x^4=\big(x^2\big)^2,~$ and $d\big(x^2\big)=2x~dx$.
A: Thank you to user170231, using the substituion $x=\sqrt{ \sin θ }$ ,we get 
$$\frac{1}{4}[\arcsin(x^2) +x^2\sqrt{1-x^4} +C] $$ as an answer. 
A: Are you allowed to use a table of integrals? 
If yes, then you may use 
$\int{\arcsin(x)\,dx}$ and 
$\int{\sqrt{1-x^2}\,dx}$. 
A: For such pattern as 
$$I=\int x^{\alpha}(a+bx^{\beta})^{\gamma}dx$$
where $\alpha,\beta,\gamma \in \mathbb{Q}$, $a,b\in \mathbb{R}$
There is a systematic procedure :
(1).If $\gamma \in \mathbb{Z}$, then $(a+bx^{\beta})^{\gamma}$ is a binomial or the inverse of a binomial. So the integral is actually in the form
$$I=\int R(x^\alpha,x^\beta)dx$$
where $R$ denotes any rational algebraic expression of its parameters. Now suppose $\alpha=\frac{n_1}{m_1},\beta=\frac{n_2}{m_2}$ in lowest term, and let $m$ be LCM of $m_1,m_2$. Let $t=x^{\frac{1}{m}}$, or $x=t^m$, and thus $dx=mt^{m-1}dt$. After this substitution, we can rewrite the integrand as a pure rational expression of $t$, i.e:
$$I=\int R(t)dt$$
Then all we have to do is just apply the systematic procedure for solving rational integrals, which is theoretically easy.
(2).If, less to our fortune (which is the general case), $\gamma \notin \mathbb{Z}$,try the substitution $t=x^\beta$,or $x=t^{\frac{1}{\beta}}$, and thus $dx=\frac{1}{\beta} t^{\frac{1}{\beta}-1}dt$, then
$$I=\frac{1}{\beta} \int t^{\frac{\alpha-\beta+1}{\beta}}(a+bt)^{\gamma}dt$$ 
Now, if you are lucky enough, then $z=\frac{\alpha-\beta+1}{\beta}$ will be an integer! In this case, apply the substitution $v=a+bt$, or $t=\frac{v-a}{b}$, and thus $dt=\frac{1}{b}dv$, then rewrite the integral as
$$I=\frac{1}{a}\int (\frac{v-b}{a})^z v^\gamma dv=\int R(v,v^\gamma)dt$$
Since $\gamma \in \mathbb Q$, we can reduce it into a pure rational expression applying the same techniques in (1).
(3).However, most probably you will come to find in case (2) that $z \notin \mathbb Z$.But don't just give up, there is still a ray of hope. As a last resort, try rewriting the integral as
$$I=\frac{1}{\beta} \int t^{\frac{\alpha}{\beta}+\frac{1}{\beta}+\gamma-1}(\frac{a+bt}{t})^{\gamma}dt$$
Now if $z'=\frac{\alpha}{\beta}+\frac{1}{\beta}+\gamma-1$ is still not integer, then it is time to give up and do never ever count on solving it (literally insoluble), but if you are lucky at last, say, if $z' \in \mathbb Z$, then congratulations! All you have to do is apply the substitution $w=(\frac{a+bt}{t})^\gamma$ and rewrite the integral, you will just find it falls within the soluble category in case (2).
