Advertisements
Advertisements
Question
\[\int \cos^{- 1} \left( 1 - 2 x^2 \right) \text{ dx }\]
Sum
Advertisements
Solution
\[\text{We have}, \]
\[I = \int \cos^{- 1} \left( 1 - 2 x^2 \right)dx\]
\[\text{ Putting x }= \sin \theta\]
\[ \Rightarrow dx = \cos \text{ θ dθ}\]
\[ \therefore I = \int \cos^{- 1} \left( 1 - 2 \sin^2 \theta \right) \cos \text{ θ dθ}\]
\[ = \int \cos^{- 1} \left( \cos 2\theta \right) \cos \text{ θ dθ}\]
\[ = 2\int \theta_I \text {cos}_{II} \text{ θ dθ}\]
\[ = 2\left[ \theta \sin \theta - \int1 \sin \text{ θ dθ}\right]\]
\[ = 2\left[ \theta \sin \theta + \cos \theta \right] + C\]
\[ = 2\left[ \sin^{- 1} x \times x + \sqrt{1 - x^2} \right] + C\]
\[ = 2\left[ x \sin^{- 1} x + \sqrt{1 - x^2} \right] + C\]
shaalaa.com
Is there an error in this question or solution?
APPEARS IN
RELATED QUESTIONS
\[\int\left( 3x\sqrt{x} + 4\sqrt{x} + 5 \right)dx\]
\[\int\frac{1}{2 - 3x} + \frac{1}{\sqrt{3x - 2}} dx\]
\[\int\frac{\left( x + 1 \right) e^x}{\cos^2 \left( x e^x \right)} dx\]
\[\int\sqrt {e^x- 1} \text{dx}\]
\[\int\frac{x^2}{\sqrt{1 - x}} dx\]
\[\int \sin^7 x \text{ dx }\]
Evaluate the following integrals:
\[\int\cos\left\{ 2 \cot^{- 1} \sqrt{\frac{1 + x}{1 - x}} \right\}dx\]
\[\int\frac{1}{\sqrt{\left( 2 - x \right)^2 + 1}} dx\]
\[\int\frac{\cos x - \sin x}{\sqrt{8 - \sin2x}}dx\]
\[\int\frac{a x^3 + bx}{x^4 + c^2} dx\]
\[\int\frac{1}{\sin^2 x + \sin 2x} \text{ dx }\]
\[\int\frac{2 \sin x + 3 \cos x}{3 \sin x + 4 \cos x} dx\]
\[\int x^3 \text{ log x dx }\]
`int"x"^"n"."log" "x" "dx"`
\[\int \sin^{- 1} \sqrt{x} \text{ dx }\]
\[\int \sin^{- 1} \left( \frac{2x}{1 + x^2} \right) \text{ dx }\]
\[\int x^2 \tan^{- 1} x\text{ dx }\]
\[\int e^x \left( \frac{x - 1}{2 x^2} \right) dx\]
\[\int e^x \frac{x - 1}{\left( x + 1 \right)^3} \text{ dx }\]
\[\int\frac{e^x}{x}\left\{ \text{ x }\left( \log x \right)^2 + 2 \log x \right\} dx\]
\[\int x^2 \sqrt{a^6 - x^6} \text{ dx}\]
\[\int\sqrt{2ax - x^2} \text{ dx}\]
\[\int\frac{5x}{\left( x + 1 \right) \left( x^2 - 4 \right)} dx\]
\[\int\frac{x^2 + 1}{x\left( x^2 - 1 \right)} dx\]
\[\int\frac{1}{x \log x \left( 2 + \log x \right)} dx\]
\[\int\frac{x^2 + 6x - 8}{x^3 - 4x} dx\]
\[\int\frac{2 x^2 + 7x - 3}{x^2 \left( 2x + 1 \right)} dx\]
\[\int\frac{2x + 1}{\left( x - 2 \right) \left( x - 3 \right)} dx\]
\[\int\frac{x^2 + 1}{x^4 + x^2 + 1} \text{ dx }\]
\[\int\frac{x^2 + 1}{x^4 + 7 x^2 + 1} 2 \text{ dx }\]
The value of \[\int\frac{\sin x + \cos x}{\sqrt{1 - \sin 2x}} dx\] is equal to
\[\int\frac{1}{\sqrt{3 - 2x - x^2}} \text{ dx}\]
\[\int\frac{1}{a + b \tan x} \text{ dx }\]
\[\int \tan^3 x\ \sec^4 x\ dx\]
\[\int e^{2x} \left( \frac{1 + \sin 2x}{1 + \cos 2x} \right) dx\]
\[\int\frac{x^2}{\left( x - 1 \right)^3 \left( x + 1 \right)} \text{ dx}\]
\[\int\sqrt{\frac{1 - \sqrt{x}}{1 + \sqrt{x}}} \text{ dx}\]
\[\int\frac{x^2}{x^2 + 7x + 10} dx\]
