Integrate the functions in Exercises: x^3 ( ^ -1 x^4 ) 1+x^8 - Vidyadip

Question

Integrate the functions in Exercises:
$\frac{\text{x}^3\sin\big(\tan^{-1}\text{x}^4\big)}{1+\text{x}^8}$

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Answer

$\text{Let I}=\int\frac{\text{x}^3\sin{(\tan^{-1}\text{x}^4)}}{1 +\text{x}^8}\text{ dx} $
$=\frac{1}{4}\int\sin(\tan^{-1}\text{x}^4)\dot\ \frac{4\text{x}^3}{1 + \text{x}^8}\text{ dx} \ \ \ \ ...\text{(i)} $
Putting $\tan^{-1} \text{x}^4=\text{t}\ \ \ $$\Rightarrow\ \ \ \frac{1}{1+(\text{x}^4)^2}\frac{\text{d}}{\text{dx}}\text{x}^4=\frac{\text{dt}}{\text{dx}}\ \ \ $$\Rightarrow \ \ \ \frac{4\text{x}^3}{1+\text{x}^8}\text{ dx = dt} $
$\therefore \ \ \ $From eq. (i), $\text{ I}=\frac{1}{4}\int\sin\text{t}\text{ dt}=\frac{-1}{4}\text{cos t}+\text{c}$
$=\frac{-1}{4}\cos(\tan^{-1}\text{x}^4)+\text{c}$

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