Integration by substitution
General
Grade 12
Question:
Find $\int \frac{\sin(\tan^{-1} x)}{1+x^2} dx$
Step-by-Step Solution
Key Concept: General
Put, $\tan^{-1} x = t \Rightarrow \frac{1}{1+x^2} dx = dt$<br>$\therefore \int \frac{\sin(\tan^{-1} x)}{1+x^2} dx = \int \sin t dt = -\cos t + C = -\cos(\tan^{-1} x) + C$
Correct Answer: $-\cos(\tan^{-1} x) + C$