Evaluate the following integrals: ( ^ -1 x) 1+x^2 dx - Vidyadip

Question

Evaluate the following integrals:
$\int\frac{\sin(\tan^{-1}\text{x})}{1+\text{x}^2}\text{ dx}$

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Answer

$\int\frac{\sin(\tan^{-1}\text{x})}{1+\text{x}^2}\text{ dx}$

Let $\tan^{-1}\text{x}=\text{t}$

$\Rightarrow\frac{1}{1+\text{x}^2}\text{ dx}=\text{dt}$

Now, $\int\frac{\sin(\tan^{-1}\text{x})}{1+\text{x}^2}\text{ dx}$

$=\int\sin\text{t dt}$

$=-\cos(\text{t})+\text{C}$

$=-\cos\big(\tan^{-1}\text{x}\big)+\text{C}$

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