मराठी

Π / 3 ∫ π / 6 1 1 + √ Cot X D X is (A) π/3 (B) π/6 (C) π/12 (D) π/2

Advertisements
Advertisements

प्रश्न

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

पर्याय

  •  π/3

  •  π/6

  • π/12

  • π/2

MCQ
Advertisements

उत्तर

\[\frac{\pi}{12}\]
\[Let\, I = \int_\frac{\pi}{6}^\frac{\pi}{3} \frac{1}{1 + \sqrt{cotx}} d x .............(1)\]
\[ = \int_\frac{\pi}{6}^\frac{\pi}{3} \frac{1}{1 + \sqrt{cot\left( \frac{\pi}{3} + \frac{\pi}{6} - x \right)}} dx ..............\left[\text{Using }\int_a^b f\left( x \right) d x = \int_a^b f\left( a + b - x \right) d x \right]\]
\[ = \int_\frac{\pi}{6}^\frac{\pi}{3} \frac{1}{1 + \sqrt{\tan x}} d x .................(2)\]
Adding (1) and (2) we get
\[2I = \int_\frac{\pi}{6}^\frac{\pi}{3} \left[ \frac{1}{1 + \sqrt{cotx}} + \frac{1}{1 + \sqrt{\tan x}} \right] d x \]
\[ = \int_\frac{\pi}{6}^\frac{\pi}{3} \frac{2 + \sqrt{cotx} + \sqrt{\tan x}}{\left( 1 + \sqrt{cotx} \right)\left( 1 + \sqrt{\tan x} \right)}dx\]
\[ = \int_\frac{\pi}{6}^\frac{\pi}{3} \left[ \frac{2 + \sqrt{cotx} + \sqrt{\tan x}}{2 + \sqrt{cotx} + \sqrt{\tan x}} \right]dx \]
\[ = \int_\frac{\pi}{6}^\frac{\pi}{3} dx\]
\[ = \left[ x \right]_\frac{\pi}{6}^\frac{\pi}{3} \]
\[ = \frac{\pi}{3} - \frac{\pi}{6}\]
\[ = \frac{\pi}{6}\]
\[\text{Hence, }I = \frac{\pi}{12}\]

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 19: Definite Integrals - MCQ [पृष्ठ ११८]

APPEARS IN

आर.डी. शर्मा Mathematics Volume 1 and 2 [English] Class 12
पाठ 19 Definite Integrals
MCQ | Q 12 | पृष्ठ ११८

संबंधित प्रश्‍न

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

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

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

Evaluate the following definite integrals:

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

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

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

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

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

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

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

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

\[\int\limits_0^{\pi/3} \frac{\cos x}{3 + 4 \sin x} dx\]

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

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

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

\[\int\limits_1^4 \left( x^2 - x \right) dx\]

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

\[\int\limits_0^{\pi/2} \log \left( \frac{3 + 5 \cos x}{3 + 5 \sin x} \right) dx .\]

 


\[\int\limits_a^b \frac{f\left( x \right)}{f\left( x \right) + f\left( a + b - x \right)} dx .\]

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

If \[\int\limits_0^a 3 x^2 dx = 8,\] write the value of a.

 

 


If \[\int_0^a \frac{1}{4 + x^2}dx = \frac{\pi}{8}\] , find the value of a.


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

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


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

The value of \[\int\limits_{- \pi}^\pi \sin^3 x \cos^2 x\ dx\] is 

 


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

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

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


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


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


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


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


\[\int\limits_{- \pi/4}^{\pi/4} \left| \tan x \right| dx\]


Using second fundamental theorem, evaluate the following:

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


Evaluate the following:

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


Verify the following:

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


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


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×