English

1 / 2 ∫ 0 1 √ 1 − X 2 D X

Advertisements
Advertisements

Question

\[\int\limits_0^{1/2} \frac{1}{\sqrt{1 - x^2}} dx\]
Advertisements

Solution

\[Let I = \int_0^\frac{1}{2} \frac{1}{\sqrt{1 - x^2}} d x . Then, \]
\[I = \left[ \sin^{- 1} x \right]_0^\frac{1}{2} \]
\[ \Rightarrow I = \sin^{- 1} \frac{1}{2} - \sin^{- 1} 0\]
\[ \Rightarrow I = \frac{\pi}{6} - 0\]
\[ \Rightarrow I = \frac{\pi}{6}\]

shaalaa.com
  Is there an error in this question or solution?
Chapter 19: Definite Integrals - Exercise 20.1 [Page 16]

APPEARS IN

R.D. Sharma Mathematics Volume 1 and 2 [English] Class 12
Chapter 19 Definite Integrals
Exercise 20.1 | Q 3 | Page 16

RELATED QUESTIONS

\[\int\limits_{- 2}^3 \frac{1}{x + 7} dx\]

\[\int\limits_0^\infty \frac{1}{a^2 + b^2 x^2} dx\]

\[\int\limits_0^\infty e^{- x} dx\]

\[\int\limits_0^{\pi/6} \cos x \cos 2x\ dx\]

\[\int\limits_{\pi/3}^{\pi/4} \left( \tan x + \cot x \right)^2 dx\]

\[\int\limits_0^{\pi/2} \cos^4\ x\ dx\]

 


\[\int\limits_0^{\pi/2} x^2 \cos\ x\ dx\]

\[\int\limits_0^{\pi/2} x^2 \cos^2 x\ dx\]

\[\int\limits_0^{2\pi} e^x \cos\left( \frac{\pi}{4} + \frac{x}{2} \right) dx\]

\[\int\limits_0^{\pi/2} \sqrt{\sin \phi} \cos^5 \phi\ d\phi\]

 


\[\int\limits_0^{\pi/2} \frac{dx}{a \cos x + b \sin x}a, b > 0\]

\[\int\limits_{- 1}^1 5 x^4 \sqrt{x^5 + 1} dx\]

\[\int\limits_0^{\pi/4} \sin^3 2t \cos 2t\ dt\]

\[\int\limits_0^{( \pi )^{2/3}} \sqrt{x} \cos^2 x^{3/2} dx\]


\[\int\limits_0^a \sin^{- 1} \sqrt{\frac{x}{a + x}} dx\]

\[\int_0^{2\pi} \cos^{- 1} \left( \cos x \right)dx\]

\[\int\limits_0^{\pi/2} \frac{1}{1 + \tan x}\]

 


\[\int\limits_0^1 \frac{\log\left( 1 + x \right)}{1 + x^2} dx\]

 


\[\int\limits_0^\pi x \sin x \cos^4 x\ dx\]

\[\int\limits_0^{\pi/2} \frac{x \sin x \cos x}{\sin^4 x + \cos^4 x} dx\]

If f (x) is a continuous function defined on [0, 2a]. Then, prove that

\[\int\limits_0^{2a} f\left( x \right) dx = \int\limits_0^a \left\{ f\left( x \right) + f\left( 2a - x \right) \right\} dx\]

 


\[\int\limits_0^2 \left( x + 3 \right) dx\]

\[\int\limits_a^b \cos\ x\ dx\]

\[\int\limits_0^2 \left( x^2 + 2 \right) dx\]

\[\int\limits_1^3 \left( 2 x^2 + 5x \right) dx\]

\[\int\limits_{- 2}^1 \frac{\left| x \right|}{x} dx .\]

\[\int\limits_2^3 \frac{1}{x}dx\]

Evaluate each of the following integral:

\[\int_0^\frac{\pi}{2} e^x \left( \sin x - \cos x \right)dx\]

 


\[\int\limits_1^2 \log_e \left[ x \right] dx .\]

\[\int\limits_{- \pi/2}^{\pi/2} \sin\left| x \right| dx\]  is equal to

\[\int\limits_0^1 \frac{x}{\left( 1 - x \right)^\frac{5}{4}} dx =\]

\[\int\limits_0^{\pi/2} \sin\ 2x\ \log\ \tan x\ dx\]  is equal to 

If f (a + b − x) = f (x), then \[\int\limits_a^b\] x f (x) dx is equal to


\[\int\limits_0^1 \tan^{- 1} x dx\]


\[\int\limits_0^1 \tan^{- 1} \left( \frac{2x}{1 - x^2} \right) dx\]


\[\int\limits_1^2 \frac{1}{x^2} e^{- 1/x} dx\]


\[\int\limits_2^3 e^{- x} dx\]


Using second fundamental theorem, evaluate the following:

`int_0^1 x"e"^(x^2)  "d"x`


Choose the correct alternative:

`int_0^1 (2x + 1)  "d"x` is


Choose the correct alternative:

`int_0^oo "e"^(-2x)  "d"x` is


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×