मराठी

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

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

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
Definite Integrals
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 20: Definite Integrals - MCQ [पृष्ठ ११८]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 12
पाठ 20 Definite Integrals
MCQ | Q 12 | पृष्ठ ११८

संबंधित प्रश्‍न

\[\int\limits_0^\infty e^{- x} dx\]

\[\int\limits_0^{\pi/4} \sec x dx\]

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

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

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

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

\[\int_0^1 x\log\left( 1 + 2x \right)dx\]

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

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

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

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

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

\[\int\limits_0^\pi \sin^3 x\left( 1 + 2 \cos x \right) \left( 1 + \cos x \right)^2 dx\]

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

\[\int_{- 2}^2 x e^\left| x \right| dx\]

\[\int_0^{2\pi} \cos^{- 1} \left( \cos x \right)dx\]

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

\[\int\limits_0^\pi \log\left( 1 - \cos x \right) dx\]

If `f` is an integrable function such that f(2a − x) = f(x), then prove that

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

 


\[\int\limits_0^2 \left( x^2 + 1 \right) dx\]

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

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

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

\[\int\limits_0^\infty e^{- x} dx .\]

Evaluate each of the following  integral:

\[\int_0^1 x e^{x^2} dx\]

 


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

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


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


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

 

\[\int\limits_{\pi/3}^{\pi/2} \frac{\sqrt{1 + \cos x}}{\left( 1 - \cos x \right)^{5/2}} dx\]


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


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


Evaluate the following using properties of definite integral:

`int_0^1 x/((1 - x)^(3/4))  "d"x`


Choose the correct alternative:

If n > 0, then Γ(n) is


Integrate `((2"a")/sqrt(x) - "b"/x^2 + 3"c"root(3)(x^2))` w.r.t. x


Evaluate `int "dx"/sqrt((x - alpha)(beta - x)), beta > alpha`


`int (cos2x - cos 2theta)/(cosx - costheta) "d"x` is equal to ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×