मराठी

Π / 2 ∫ 0 1 2 + Cos X D X Equals

Advertisements
Advertisements

प्रश्न

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

पर्याय

  • \[\frac{1}{3} \tan^{- 1} \left( \frac{1}{\sqrt{3}} \right)\]
  • \[\frac{2}{\sqrt{3}} \tan^{- 1} \left( \frac{1}{\sqrt{3}} \right)\]
  • \[\sqrt{3} \tan^{- 1} \left( \sqrt{3} \right)\]

     

  • \[2\sqrt{3} \tan^{- 1} \sqrt{3}\]
MCQ
Advertisements

उत्तर

\[\ \frac{2}{\sqrt{3}} \tan^{- 1} \left( \frac{1}{\sqrt{3}} \right)\]

\[\text{We have}, \]

\[I = \int_0^\frac{\pi}{2} \frac{1}{2 + \cos x} d x\]

\[ = \int_0^\frac{\pi}{2} \frac{1}{2 + \frac{1 - \tan^2 \frac{x}{2}}{1 + \tan^2 \frac{x}{2}}} d x\]

\[ = \int_0^\frac{\pi}{2} \frac{1 + \tan^2 \frac{x}{2}}{2 + 2 \tan^2 \frac{x}{2} + 1 - \tan^2 \frac{x}{2}}dx\]

\[ = \int_0^\frac{\pi}{2} \frac{\sec^2 \frac{x}{2}}{3 + \tan^2 \frac{x}{2}}dx\]

\[\text{Putting} \tan \frac{x}{2} = t\]

\[ \Rightarrow \frac{1}{2} \sec^2 \frac{x}{2}dx = dt\]

\[ \Rightarrow \sec^2 \frac{x}{2}dx = 2dt\]

\[When, x \to 0; t \to 0\]

\[and x \to \frac{\pi}{2}; t \to 1\]

\[ \therefore I = \int_0^1 \frac{2}{3 + t^2}dt\]

\[ = 2 \int_0^1 \frac{1}{\left( \sqrt{3} \right)^2 + t^2}dt\]

\[ = \frac{2}{\sqrt{3}} \left[ \tan^{- 1} \frac{t}{\sqrt{3}} \right]_0^1 \]

\[ = \frac{2}{\sqrt{3}}\left[ \tan^{- 1} \frac{1}{\sqrt{3}} - \tan^{- 1} \frac{0}{\sqrt{3}} \right]\]

\[ = \frac{2}{\sqrt{3}} \tan^{- 1} \left( \frac{1}{\sqrt{3}} \right)\]

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 9 | पृष्ठ ११७

संबंधित प्रश्‍न

\[\int\limits_{\pi/4}^{\pi/2} \cot x\ dx\]


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

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

\[\int_0^\frac{\pi}{4} \left( a^2 \cos^2 x + b^2 \sin^2 x \right)dx\]

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

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

\[\int\limits_0^{\pi/2} \cos^5 x\ dx\]

\[\int\limits_4^9 \frac{\sqrt{x}}{\left( 30 - x^{3/2} \right)^2} dx\]

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

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

\[\int\limits_0^9 f\left( x \right) dx, where f\left( x \right) \begin{cases}\sin x & , & 0 \leq x \leq \pi/2 \\ 1 & , & \pi/2 \leq x \leq 3 \\ e^{x - 3} & , & 3 \leq x \leq 9\end{cases}\]

\[\int_{- \frac{\pi}{2}}^\frac{\pi}{2} \frac{- \frac{\pi}{2}}{\sqrt{\cos x \sin^2 x}}dx\]

\[\int_0^2 2x\left[ x \right]dx\]

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

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

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

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

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

\[\int\limits_0^4 \frac{1}{\sqrt{16 - x^2}} dx .\]

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

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

Evaluate : 

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

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

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

 


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 value of \[\int\limits_0^\pi \frac{1}{5 + 3 \cos x} dx\] is

 


\[\int\limits_0^{2a} f\left( x \right) dx\]  is equal to


Evaluate : \[\int e^{2x} \cdot \sin \left( 3x + 1 \right) dx\] .


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


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


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


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


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


\[\int\limits_{- a}^a \frac{x e^{x^2}}{1 + x^2} dx\]


Evaluate the following using properties of definite integral:

`int_0^1 x/((1 - x)^(3/4))  "d"x`


Evaluate the following:

`int_0^oo "e"^(-mx) x^6 "d"x`


Choose the correct alternative:

If f(x) is a continuous function and a < c < b, then `int_"a"^"c" f(x)  "d"x + int_"c"^"b" f(x)  "d"x` is


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


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×