English

∫ 1 √ 2 X − X 2 D X - Mathematics

Advertisements
Advertisements

Question

\[\int\frac{1}{\sqrt{2x - x^2}} dx\]
Sum
Advertisements

Solution

\[\int\frac{dx}{\sqrt{2x - x^2}}\]
\[ = \int\frac{dx}{\sqrt{2x - x^2 - 1 + 1}}\]
\[ = \int\frac{dx}{\sqrt{1 - \left( x^2 - 2x + 1 \right)}}\]
\[ = \int\frac{dx}{\sqrt{1 - \left( x - 1 \right)^2}} \]
\[ = \sin^{- 1} \left( x - 1 \right) + C \left[ \because \int\frac{dx}{\sqrt{a^2 - x^2}} = \sin^{- 1} \left( \frac{x}{a} \right) + C \right]\]

shaalaa.com
  Is there an error in this question or solution?
Chapter 19: Indefinite Integrals - Exercise 19.17 [Page 93]

APPEARS IN

RD Sharma Mathematics [English] Class 12
Chapter 19 Indefinite Integrals
Exercise 19.17 | Q 1 | Page 93

Video TutorialsVIEW ALL [1]

RELATED QUESTIONS

\[\int\frac{2 x^4 + 7 x^3 + 6 x^2}{x^2 + 2x} dx\]

\[\int\frac{1}{1 - \sin x} dx\]

\[\int\sin x\sqrt{1 + \cos 2x} dx\]

\[\int\frac{1 - \cos x}{1 + \cos x} dx\]

\[\int \left( e^x + 1 \right)^2 e^x dx\]

`∫     cos ^4  2x   dx `


\[\int \sin^2\text{ b x dx}\]

Integrate the following integrals:

\[\int\text{sin 2x  sin 4x    sin 6x  dx} \]

\[\int \sin^5\text{ x }\text{cos x dx}\]

\[\int\frac{\sin 2x}{\left( a + b \cos 2x \right)^2} dx\]

\[\int\frac{\cos\sqrt{x}}{\sqrt{x}} dx\]

\[\int \tan^3 \text{2x sec 2x dx}\]

\[\int\sqrt {e^x- 1}  \text{dx}\] 

\[\int\frac{x^5}{\sqrt{1 + x^3}} dx\]

` ∫   tan   x   sec^4  x   dx  `


\[\int\frac{x - 1}{3 x^2 - 4x + 3} dx\]

\[\int\frac{\sin 2x}{\sin^4 x + \cos^4 x} \text{ dx }\]

\[\int\frac{1}{\cos 2x + 3 \sin^2 x} dx\]

`int 1/(cos x - sin x)dx`

\[\int\frac{5 \cos x + 6}{2 \cos x + \sin x + 3} \text{ dx }\]

\[\int\frac{\log \left( \log x \right)}{x} dx\]

\[\int x^2 \sin^2 x\ dx\]

\[\int x^3 \cos x^2 dx\]

\[\int e^x \frac{1 + x}{\left( 2 + x \right)^2} \text{ dx }\]

\[\int\left( x + 1 \right) \sqrt{x^2 + x + 1} \text{  dx }\]

\[\int\frac{3 + 4x - x^2}{\left( x + 2 \right) \left( x - 1 \right)} dx\]

\[\int\left( x - 1 \right) e^{- x} dx\] is equal to

\[\int\frac{x + 2}{\left( x + 1 \right)^3} \text{ dx }\]


\[\int\frac{x^4 + x^2 - 1}{x^2 + 1} \text{ dx}\]

\[\int\frac{\sin x}{\cos 2x} \text{ dx }\]

\[\int x \sin^5 x^2 \cos x^2 dx\]

\[\int\sqrt{\sin x} \cos^3 x\ \text{ dx }\]

\[\int\frac{\sqrt{a} - \sqrt{x}}{1 - \sqrt{ax}}\text{  dx }\]

\[\int\frac{1}{\sin^2 x + \sin 2x} \text{ dx }\]

\[\int\frac{x^3}{\sqrt{x^8 + 4}} \text{ dx }\]


\[\int\frac{1}{2 - 3 \cos 2x} \text{ dx }\]


\[\int\frac{1}{\sin^4 x + \cos^4 x} \text{ dx}\]


\[\int x\sqrt{\frac{1 - x}{1 + x}} \text{ dx }\]

\[\int\frac{1}{x\sqrt{1 + x^3}} \text{ dx}\]

Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×