Question
Integrate the function w.r.t. x: $\frac{\sin \left(\tan ^{-1} x\right)}{1+x^{2}}$
✓
Answer
Derivative of $\tan ^{-1} x=\frac{1}{1+x^{2}}$. Thus, we use the substitution $\tan^{-1} x = t$ so that $\frac{d x}{1+x^{2}}=d t$
Therefore, $\int \frac{\sin \left(\tan ^{-1} x\right)}{1+x^{2}} d x=\int \sin t d t = -\cos t + C = -\cos(\tan^{-1}x) + C$
Therefore, $\int \frac{\sin \left(\tan ^{-1} x\right)}{1+x^{2}} d x=\int \sin t d t = -\cos t + C = -\cos(\tan^{-1}x) + C$
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