Question

Find \(\int \dfrac{x}{\sqrt{1-x^4}}\) Please, help

Answer 1

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Evaluate the indefinite integral:

$\mathsf{\displaystyle\int\! \frac{x}{\sqrt{1-x^4}}\,dx}\\\\\\ \mathsf{=\displaystyle\int\! \frac{1}{2}\cdot 2\cdot \frac{1}{\sqrt{1-(x^2)^2}}\,dx}\\\\\\ \mathsf{=\displaystyle \frac{1}{2}\int\! \frac{1}{\sqrt{1-(x^2)^2}}\cdot 2x\,dx\qquad\quad(i)}$


Make a trigonometric substitution:

$\begin{array}{lcl} \mathsf{x^2=sin\,t}&\quad\Rightarrow\quad&\mathsf{2x\,dx=cos\,t\,dt}\\\\ &&\mathsf{t=arcsin(x^2)\,,\qquad 0\ \textless \ x\ \textless \ \frac{\pi}{2}}\end{array}$


so the integral (i) becomes

$\mathsf{=\displaystyle\frac{1}{2}\int\!\frac{1}{\sqrt{1-sin^2\,t}}\cdot cos\,t\,dt\qquad\quad (but~1-sin^2\,t=cos^2\,t)}\\\\\\ \mathsf{=\displaystyle\frac{1}{2}\int\!\frac{1}{\sqrt{cos^2\,t}}\cdot cos\,t\,dt}$

$\mathsf{=\displaystyle\frac{1}{2}\int\!\frac{1}{cos\,t}\cdot cos\,t\,dt}\\\\\\ \mathsf{=\displaystyle\frac{1}{2}\int\!\f dt}\\\\\\ \mathsf{=\displaystyle\frac{1}{2}\,t+C}$


Now, substitute back for t = arcsin(x²), and you finally get the result:

$\mathsf{\displaystyle\int\! \frac{x}{\sqrt{1-(x^2)^2}}\,dx=\frac{1}{2}\,arcsin(x^2)+C}$          ✔

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You could also make

x² = cos t

and you would get this expression for the integral:

$\mathsf{\displaystyle\int\! \frac{x}{\sqrt{1-(x^2)^2}}\,dx=-\,\frac{1}{2}\,arccos(x^2)+C_2}$          ✔


which is fine, because those two functions have the same derivative, as the difference between them is a constant:

$\mathsf{\dfrac{1}{2}\,arcsin(x^2)-\left(-\dfrac{1}{2}\,arccos(x^2)\right)}\\\\\\ =\mathsf{\dfrac{1}{2}\,arcsin(x^2)+\dfrac{1}{2}\,arccos(x^2)}\\\\\\ =\mathsf{\dfrac{1}{2}\cdot \left[\,arcsin(x^2)+arccos(x^2)\right]}\\\\\\ =\mathsf{\dfrac{1}{2}\cdot \dfrac{\pi}{2}}$

$\mathsf{=\dfrac{\pi}{4}}$         ✔


and that constant does not interfer in the differentiation process, because the derivative of a constant is zero.


I hope this helps. =)