English

Π / 2 ∫ 0 √ 1 − Cos 2 X D X . - Mathematics

Advertisements
Advertisements

Question

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

Solution

\[\int_0^\frac{\pi}{2} \sqrt{1 - \cos2x}\ dx\]

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

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

\[ = - \sqrt{2} \left[ \cos x \right]_0^\frac{\pi}{2} \]

\[ = - \left( 0 - \sqrt{2} \right)\]

\[ = \sqrt{2}\]

\[\]

shaalaa.com
Definite Integrals
  Is there an error in this question or solution?
Chapter 20: Definite Integrals - Very Short Answers [Page 115]

APPEARS IN

RD Sharma Mathematics [English] Class 12
Chapter 20 Definite Integrals
Very Short Answers | Q 13 | Page 115

RELATED QUESTIONS

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

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

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

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

\[\int\limits_0^{\pi/2} \sin^3 x\ dx\]

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

\[\int\limits_1^3 \frac{\cos \left( \log x \right)}{x} dx\]

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

\[\int\limits_0^\pi \frac{1}{3 + 2 \sin x + \cos x} dx\]

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

\[\int\limits_0^\pi 5 \left( 5 - 4 \cos \theta \right)^{1/4} \sin \theta\ d \theta\]

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

Evaluate each of the following integral:

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


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

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

\[\int\limits_{- \pi/2}^{\pi/2} \sin^3 x\ dx\]

\[\int\limits_{- \pi/2}^{\pi/2} \log\left( \frac{2 - \sin x}{2 + \sin x} \right) dx\]

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

\[\int\limits_3^5 \left( 2 - x \right) dx\]

\[\int\limits_0^2 \left( x^2 + 1 \right) dx\]

\[\int\limits_0^2 \left( x^2 + 4 \right) dx\]

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

Evaluate each of the following integral:

\[\int_0^\frac{\pi}{4} \tan\ xdx\]

 


Evaluate each of the following integral:

\[\int_0^\frac{\pi}{4} \sin2xdx\]

Evaluate each of the following integral:

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

 


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

\[\int\limits_0^\pi \frac{1}{1 + \sin x} dx\] equals


Given that \[\int\limits_0^\infty \frac{x^2}{\left( x^2 + a^2 \right)\left( x^2 + b^2 \right)\left( x^2 + c^2 \right)} dx = \frac{\pi}{2\left( a + b \right)\left( b + c \right)\left( c + a \right)},\] the value of \[\int\limits_0^\infty \frac{dx}{\left( x^2 + 4 \right)\left( x^2 + 9 \right)},\]


If \[\int\limits_0^a \frac{1}{1 + 4 x^2} dx = \frac{\pi}{8},\] then a equals

 


The derivative of \[f\left( x \right) = \int\limits_{x^2}^{x^3} \frac{1}{\log_e t} dt, \left( x > 0 \right),\] is

 


If \[I_{10} = \int\limits_0^{\pi/2} x^{10} \sin x\ dx,\]  then the value of I10 + 90I8 is

 


\[\int\limits_0^4 x\sqrt{4 - x} dx\]


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


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


\[\int\limits_{- 1/2}^{1/2} \cos x \log\left( \frac{1 + x}{1 - x} \right) dx\]


\[\int\limits_0^\pi \cos 2x \log \sin x dx\]


Evaluate the following:

Γ(4)


Choose the correct alternative:

`Γ(3/2)`


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×