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# Integrate the function$\int\frac{x^3}{\sqrt{1-x^8}}$

$\begin{array}{1 1} \large \frac{1}{4} \sin^{-1}(x^4)+c \\ \large \frac{1}{4} \cos^{-1}(x^4)+c \\ \large \frac{1}{4} \sin(x^4)+c \\ \large \frac{1}{4} \cos (x^4)+c\end{array}$

Can you answer this question?

Toolbox:
• (i) If in a function $\int f(x)dx,\;let\;f(x)=t,\;then\;f'(x)=dt,\;then \;\int f(x)dx=\int tdt$
• (ii)$\int \large\frac{dx}{\sqrt {a^2-x^2}}=\sin^{-1}(x/a)+c$
given $I=\large\int\frac{x^3}{\sqrt{1-x^8}}dx$

Let $x^4=t;$ on differentiating we get

$4x^3dx=dt$

$=>x^3dx=\frac{dt}{4}$

Hence $I=\int \frac{dt/4}{\sqrt {1-t^2}}=\frac{1}{4}\int \frac{dt}{\sqrt{1-t^2}}$

On integrating we get, since it is of the form $\int \frac{dx}{\sqrt {a^2-x^2}}$

$I=\frac{1}{4}\int \frac{dt}{\sqrt{1-t^2}}=\sin^{-1}(t)+c$

Substituting for t we get

$I=\frac{1}{4} \sin^{-1}(x^4)+c$

answered Feb 17, 2013 by