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Integrate the function
$\int \frac{x^3}{\sqrt{1-x^8}} d x=1$ (say)
$\int \frac{x^3}{\sqrt{1-x^8}} d x=1$ (say)
Solution:
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Put $\mathrm{x}^4=\mathrm{t}$ so that $4 \mathrm{x}^3 \mathrm{dx}=\mathrm{dt}$,
$\mathrm{I}=\frac{1}{4} \int \frac{\mathrm{dt}}{\sqrt{1-\mathrm{t}^2}}=\frac{1}{4} \sin ^{-1} \mathrm{t}+\mathrm{c},=\frac{1}{4} \sin ^{-1} \mathrm{x}^4+\mathrm{c}$
$\mathrm{I}=\frac{1}{4} \int \frac{\mathrm{dt}}{\sqrt{1-\mathrm{t}^2}}=\frac{1}{4} \sin ^{-1} \mathrm{t}+\mathrm{c},=\frac{1}{4} \sin ^{-1} \mathrm{x}^4+\mathrm{c}$
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