Advertisements
Advertisements
Question
Solution
\[\text{We have}, \]
\[I = \int \cos^{- 1} \left( 1 - 2 x^2 \right)dx\]
\[\text{ Putting x }= \sin \theta\]
\[ \Rightarrow dx = \cos \text{ θ dθ}\]
\[ \therefore I = \int \cos^{- 1} \left( 1 - 2 \sin^2 \theta \right) \cos \text{ θ dθ}\]
\[ = \int \cos^{- 1} \left( \cos 2\theta \right) \cos \text{ θ dθ}\]
\[ = 2\int \theta_I \text {cos}_{II} \text{ θ dθ}\]
\[ = 2\left[ \theta \sin \theta - \int1 \sin \text{ θ dθ}\right]\]
\[ = 2\left[ \theta \sin \theta + \cos \theta \right] + C\]
\[ = 2\left[ \sin^{- 1} x \times x + \sqrt{1 - x^2} \right] + C\]
\[ = 2\left[ x \sin^{- 1} x + \sqrt{1 - x^2} \right] + C\]
APPEARS IN
RELATED QUESTIONS
If f' (x) = x − \[\frac{1}{x^2}\] and f (1) \[\frac{1}{2}, find f(x)\]
If \[\int\frac{\sin^8 x - \cos^8 x}{1 - 2 \sin^2 x \cos^2 x} dx\]