English

The Value of the Integral π ∫ 0 / 2 √ Cos X √ Cos X + √ Sin X D X is

Advertisements
Advertisements

Question

The value of the integral \[\int\limits_0^{\pi/2} \frac{\sqrt{\cos x}}{\sqrt{\cos x} + \sqrt{\sin x}} dx\]  is 

Options

  • 0

  • π/2

  • π/4

  • none of these

MCQ
Advertisements

Solution

π/4

 

\[Let\ I = \int_0^\frac{\pi}{2} \frac{\sqrt{\cos x}}{\sqrt{\cos x} + \sqrt{\sin x}} d x ..............(1)\]
\[ = \int_0^\frac{\pi}{2} \frac{\sqrt{\cos\left( \frac{\pi}{2} - x \right)}}{\sqrt{\cos\left( \frac{\pi}{2} - x \right)} + \sqrt{\sin\left( \frac{\pi}{2} - x \right)}} d x \]
\[ = \int_0^\frac{\pi}{2} \frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}}dx\]
\[ = \int_0^\frac{\pi}{2} \frac{\sqrt{\sin x}}{\sqrt{\cos x} + \sqrt{\sin x}}dx ...............(2)\]
\[\text{Adding (1) and (2)}\]
\[2I = \int_0^\frac{\pi}{2} \left[ \frac{\sqrt{\cos x}}{\sqrt{\cos x} + \sqrt{\sin x}} + \frac{\sqrt{\sin x}}{\sqrt{\cos x} + \sqrt{\sin x}} \right] d x\]
\[ = \int_0^\frac{\pi}{2} dx\]
\[ = \left[ x \right]_0^\frac{\pi}{2} = \frac{\pi}{2}\]
\[Hence\ I = \frac{\pi}{4}\]

shaalaa.com
  Is there an error in this question or solution?
Chapter 19: Definite Integrals - MCQ [Page 117]

APPEARS IN

R.D. Sharma Mathematics Volume 1 and 2 [English] Class 12
Chapter 19 Definite Integrals
MCQ | Q 5 | Page 117

RELATED QUESTIONS

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

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

\[\int\limits_1^3 \frac{\log x}{\left( x + 1 \right)^2} dx\]

\[\int\limits_1^4 \frac{x^2 + x}{\sqrt{2x + 1}} dx\]

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

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

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

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

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

\[\int_{- \frac{\pi}{4}}^\frac{\pi}{2} \sin x\left| \sin x \right|dx\]

 


\[\int\limits_0^\pi x \sin^3 x\ dx\]

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

Evaluate the following integral:

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

 


\[\int_0^1 | x\sin \pi x | dx\]

\[\int\limits_0^2 e^x dx\]

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

\[\int\limits_0^{\pi/2} \log \tan x\ dx .\]

Evaluate each of the following integral:

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

\[\int\limits_0^3 \frac{3x + 1}{x^2 + 9} dx =\]

\[\lim_{n \to \infty} \left\{ \frac{1}{2n + 1} + \frac{1}{2n + 2} + . . . + \frac{1}{2n + n} \right\}\] is equal to

Evaluate : \[\int\limits_0^\pi/4 \frac{\sin x + \cos x}{16 + 9 \sin 2x}dx\] .


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


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


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


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


\[\int\limits_{\pi/6}^{\pi/2} \frac{\ cosec x \cot x}{1 + {cosec}^2 x} dx\]


Evaluate the following:

`int_(-1)^1 "f"(x)  "d"x` where f(x) = `{{:(x",", x ≥ 0),(-x",", x  < 0):}`


Evaluate the following using properties of definite integral:

`int_(- pi/2)^(pi/2) sin^2theta  "d"theta`


Choose the correct alternative:

`int_0^oo "e"^(-2x)  "d"x` is


Choose the correct alternative:

Γ(n) is


Find `int x^2/(x^4 + 3x^2 + 2) "d"x`


Evaluate `int (x^2"d"x)/(x^4 + x^2 - 2)`


Verify the following:

`int (x - 1)/(2x + 3) "d"x = x - log |(2x + 3)^2| + "C"`


Evaluate the following:

`int ((x^2 + 2))/(x + 1) "d"x`


`int (x + 3)/(x + 4)^2 "e"^x  "d"x` = ______.


What is the meaning of an indefinite integral?


Using \[F\] as an antiderivative, how is \[\int_{a}^{b}f(x)\,dx\] evaluated?


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×