मराठी

Π / 2 ∫ 0 1 1 + Cot 3 X D X is Equal to (A) 0 (B) 1 (C) π/2 (D) π/4 - Mathematics

Advertisements
Advertisements

प्रश्न

\[\int\limits_0^{\pi/2} \frac{1}{1 + \cot^3 x} dx\]  is equal to

पर्याय

  • 0

  • 1

  • π/2

  • π/4

MCQ
Advertisements

उत्तर

 π/4

 

We have
\[ I = \int_0^\frac{\pi}{2} \frac{1}{1 + \cot^3 x} d x . . . . . \left( 1 \right)\]
\[ = \int_0^\frac{\pi}{2} \frac{1}{1 + \cot^3 \left( \frac{\pi}{2} - x \right)} d x \]
\[ \therefore I = \int_0^\frac{\pi}{2} \frac{1}{1 + \tan^3 x} d x . . . . . \left( 2 \right)\]
\[\text{Adding} \left( 1 \right) and \left( 2 \right) \text{we get}\]
\[2I = \int_0^\frac{\pi}{2} \left[ \frac{1}{1 + co t^3 x} + \frac{1}{1 + \tan^3 x} \right] d x\]

\[= \int_0^\frac{\pi}{2} \left[ \frac{1 + \tan^3 x + 1 + co t^3 x}{\left( 1 + co t^3 x \right)\left( 1 + \tan^3 x \right)} \right] dx\]
\[ = \int_0^\frac{\pi}{2} \left[ \frac{2 + \tan^3 x + co t^3 x}{1 + \tan^3 x + co t^3 x + co t^3 x \tan^3 x} \right]dx\]
\[ = \int_0^\frac{\pi}{2} \left[ \frac{2 + \tan^3 x + co t^3 x}{1 + \tan^3 x + co t^3 x + 1} \right]dx\]
\[ = \int_0^\frac{\pi}{2} \left[ \frac{2 + \tan^3 x + co t^3 x}{2 + \tan^3 x + co t^3 x} \right] dx\]
\[ = \int_0^\frac{\pi}{2} [1]dx\]
\[ = \left[ x \right]_0^\frac{\pi}{2} \]
\[ = \frac{\pi}{2}\]
\[Hence\ I = \frac{\pi}{4}\]

shaalaa.com
Definite Integrals
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 20: Definite Integrals - MCQ [पृष्ठ ११९]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 12
पाठ 20 Definite Integrals
MCQ | Q 31 | पृष्ठ ११९

संबंधित प्रश्‍न

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

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

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

\[\int\limits_0^{\pi/2} \frac{\sin x \cos x}{1 + \sin^4 x} dx\]

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

\[\int\limits_0^5 \frac{\sqrt[4]{x + 4}}{\sqrt[4]{x + 4} + \sqrt[4]{9 - x}} dx\]

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

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

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

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

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

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

 


\[\int\limits_0^2 \left( 3 x^2 - 2 \right) dx\]

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

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

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

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

If \[f\left( x \right) = \int_0^x t\sin tdt\], the write the value of \[f'\left( x \right)\]


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

\[\int\limits_0^1 e^\left\{ x \right\} dx .\]

\[\int\limits_0^\pi \frac{1}{a + b \cos x} dx =\]

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

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

 


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

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

 

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


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


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


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


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


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


Evaluate the following using properties of definite integral:

`int_(- pi/2)^(pi/2) sin^2theta  "d"theta`


Evaluate the following using properties of definite integral:

`int_0^1 log (1/x - 1)  "d"x`


Evaluate the following integrals as the limit of the sum:

`int_1^3 (2x + 3)  "d"x`


Choose the correct alternative:

Using the factorial representation of the gamma function, which of the following is the solution for the gamma function Γ(n) when n = 8 is


Choose the correct alternative:

Γ(1) is


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


`int x^9/(4x^2 + 1)^6  "d"x` is equal to ______.


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


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×