मराठी

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

Advertisements
Advertisements

प्रश्न

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

पर्याय

  • 0

  • π/2

  • π/4

  • none of these

MCQ
Advertisements

उत्तर

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
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 19: Definite Integrals - MCQ [पृष्ठ ११७]

APPEARS IN

आर.डी. शर्मा Mathematics Volume 1 and 2 [English] Class 12
पाठ 19 Definite Integrals
MCQ | Q 5 | पृष्ठ ११७

संबंधित प्रश्‍न

\[\int\limits_4^9 \frac{1}{\sqrt{x}} dx\]

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

\[\int\limits_{\pi/6}^{\pi/4} cosec\ x\ dx\]

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

\[\int\limits_0^{\pi/2} \left( a^2 \cos^2 x + b^2 \sin^2 x \right) dx\]

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

\[\int\limits_0^1 x \left( 1 - x \right)^5 dx\]

\[\int_\frac{\pi}{6}^\frac{\pi}{3} \left( \tan x + \cot x \right)^2 dx\]

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

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

\[\int\limits_0^{\pi/4} \left( \sqrt{\tan}x + \sqrt{\cot}x \right) dx\]

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

\[\int_0^\frac{\pi}{4} \frac{\sin x + \cos x}{3 + \sin2x}dx\]

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

\[\int\limits_{- a}^a \sqrt{\frac{a - x}{a + x}} dx\]

\[\int_{- 1}^2 \left( \left| x + 1 \right| + \left| x \right| + \left| x - 1 \right| \right)dx\]

 


\[\int_{- \frac{\pi}{2}}^\pi \sin^{- 1} \left( \sin x \right)dx\]

Evaluate each of the following integral:

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


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

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

If f is an integrable function, show that

\[\int\limits_{- a}^a x f\left( x^2 \right) dx = 0\]

 


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

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

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

Evaluate each of the following integral:

\[\int_e^{e^2} \frac{1}{x\log x}dx\]

Evaluate : 

\[\int\limits_2^3 3^x dx .\]

\[\int\limits_0^1 \left\{ x \right\} dx,\] where {x} denotes the fractional part of x.  

 

\[\int\limits_{\pi/6}^{\pi/3} \frac{1}{1 + \sqrt{\cot}x} dx\] is

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 \[\int\limits_0^\infty \frac{dx}{\left( x^2 + 4 \right)\left( x^2 + 9 \right)},\]


The value of \[\int\limits_0^1 \tan^{- 1} \left( \frac{2x - 1}{1 + x - x^2} \right) dx,\] is


The value of \[\int\limits_0^{\pi/2} \log\left( \frac{4 + 3 \sin x}{4 + 3 \cos x} \right) dx\] is 

 


Evaluate: \[\int\limits_{- \pi/2}^{\pi/2} \frac{\cos x}{1 + e^x}dx\] .

 

\[\int\limits_0^{\pi/4} \sin 2x \sin 3x dx\]


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


\[\int\limits_0^{\pi/2} \left| \sin x - \cos x \right| dx\]


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


Choose the correct alternative:

`int_0^oo x^4"e"^-x  "d"x` is


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


The value of `int_2^3 x/(x^2 + 1)`dx is ______.


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


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×