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

\[\int\limits_0^{\pi/2} \sqrt{1 + \cos x}\ dx\]

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

\[\int\limits_0^1 \frac{2x + 3}{5 x^2 + 1} dx\]

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

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

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

\[\int_0^\frac{\pi}{4} \left( \tan x + \cot x \right)^{- 2} dx\]

\[\int_0^1 x\log\left( 1 + 2x \right)dx\]

\[\int\limits_0^a \frac{x}{\sqrt{a^2 + x^2}} dx\]

\[\int\limits_0^1 \frac{2x}{1 + x^4} dx\]

\[\int\limits_0^a \sqrt{a^2 - x^2} dx\]

\[\int\limits_0^{\pi/2} \frac{\sin \theta}{\sqrt{1 + \cos \theta}} d\theta\]

\[\int\limits_0^{\pi/2} 2 \sin x \cos x \tan^{- 1} \left( \sin x \right) dx\]

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

Evaluate the following integral:

\[\int\limits_{- 2}^2 \left| 2x + 3 \right| dx\]

\[\int_{- 2}^2 x e^\left| x \right| dx\]

Evaluate each of the following integral:

\[\int_a^b \frac{x^\frac{1}{n}}{x^\frac{1}{n} + \left( a + b - x \right)^\frac{1}{n}}dx, n \in N, n \geq 2\]


\[\int\limits_0^a \frac{1}{x + \sqrt{a^2 - x^2}} dx\]

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

Evaluate the following integral:

\[\int_{- 1}^1 \left| xcos\pi x \right|dx\]

 


\[\int\limits_0^3 \left( x + 4 \right) dx\]

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

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

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

If \[\left[ \cdot \right] and \left\{ \cdot \right\}\] denote respectively the greatest integer and fractional part functions respectively, evaluate the following integrals:

\[\int\limits_0^{\pi/4} \sin \left\{ x \right\} dx\]

 


\[\int_0^\frac{\pi^2}{4} \frac{\sin\sqrt{x}}{\sqrt{x}} dx\] equals


The value of \[\int\limits_0^{\pi/2} \cos x\ e^{\sin x}\ dx\] is

 


The value of \[\int\limits_{- \pi}^\pi \sin^3 x \cos^2 x\ dx\] is 

 


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

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


The value of \[\int\limits_{- \pi/2}^{\pi/2} \left( x^3 + x \cos x + \tan^5 x + 1 \right) dx, \] is 


Evaluate : \[\int\frac{dx}{\sin^2 x \cos^2 x}\] .


\[\int\limits_0^\pi \sin^3 x\left( 1 + 2 \cos x \right) \left( 1 + \cos x \right)^2 dx\]


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


\[\int\limits_1^3 \left| x^2 - 4 \right| dx\]


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


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


Evaluate the following using properties of definite integral:

`int_0^1 x/((1 - x)^(3/4))  "d"x`


Evaluate the following integrals as the limit of the sum:

`int_0^1 x^2  "d"x`


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×