मराठी

Π / 3 ∫ π / 6 1 Sin 2 X D X is Equal to , Loge 3 ,Log E √ 3,1 2 Log ( − 1 ), Log (−1)

Advertisements
Advertisements

प्रश्न

\[\int\limits_{\pi/6}^{\pi/3} \frac{1}{\sin 2x} dx\]  is equal to

पर्याय

  •  loge 3

  • \[\log_e \sqrt{3}\]
  • \[\frac{1}{2}\log\left( - 1 \right)\]
  •  log (−1)

     
MCQ
Advertisements

उत्तर

\[\log_e \sqrt{3}\]

\[\int_\frac{\pi}{6}^\frac{\pi}{3} \frac{1}{\sin2x} d x\]
\[ = \int_\frac{\pi}{6}^\frac{\pi}{3} \ cosec2x\ dx\]
\[ = \frac{1}{2} \int_\frac{\pi}{6}^\frac{\pi}{3} 2\ cosec2x\ dx\]
\[ = \frac{- 1}{2} \left[ \log\left( \ cosec\ 2x\ + \cot2x \right) \right]_\frac{\pi}{6}^\frac{\pi}{3} \]
\[ = \frac{- 1}{2}\left[ - 2\log\sqrt{3} \right]\]
\[ = \log\sqrt{3}\]

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 19: Definite Integrals - MCQ [पृष्ठ ११९]

APPEARS IN

आर.डी. शर्मा Mathematics Volume 1 and 2 [English] Class 12
पाठ 19 Definite Integrals
MCQ | Q 24 | पृष्ठ ११९

संबंधित प्रश्‍न

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

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

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

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

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

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

\[\int\limits_0^{\pi/2} \sqrt{\sin \phi} \cos^5 \phi\ d\phi\]

 


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

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

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

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

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

\[\int\limits_1^4 f\left( x \right) dx, where f\left( x \right) = \begin{cases}7x + 3 & , & \text{if }1 \leq x \leq 3 \\ 8x & , & \text{if }3 \leq x \leq 4\end{cases}\]


Evaluate the following integral:

\[\int\limits_{- 3}^3 \left| x + 1 \right| dx\]

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

 


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

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

If f(2a − x) = −f(x), prove that

\[\int\limits_0^{2a} f\left( x \right) dx = 0 .\]

If f is an integrable function, show that

\[\int\limits_{- a}^a f\left( x^2 \right) dx = 2 \int\limits_0^a f\left( x^2 \right) dx\]


\[\int\limits_{- 1}^1 \left( x + 3 \right) dx\]

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

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

\[\int\limits_1^2 \log_e \left[ x \right] dx .\]

\[\int\limits_0^\sqrt{2} \left[ x^2 \right] dx .\]

`int_0^1 sqrt((1 - "x")/(1 + "x")) "dx"`


The value of the integral \[\int\limits_{- 2}^2 \left| 1 - x^2 \right| dx\] is ________ .


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


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


Evaluate : \[\int\limits_0^\pi \frac{x}{1 + \sin \alpha \sin x}dx\] .


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


\[\int\limits_0^{1/\sqrt{3}} \tan^{- 1} \left( \frac{3x - x^3}{1 - 3 x^2} \right) dx\]


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


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


Using second fundamental theorem, evaluate the following:

`int_0^3 ("e"^x "d"x)/(1 + "e"^x)`


Evaluate the following:

`int_1^4` f(x) dx where f(x) = `{{:(4x + 3",", 1 ≤ x ≤ 2),(3x + 5",", 2 < x ≤ 4):}`


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


If `intx^3/sqrt(1 + x^2) "d"x = "a"(1 + x^2)^(3/2) + "b"sqrt(1 + x^2) + "C"`, then ______.


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


Geometrically, a definite integral is interpreted as what?


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×