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