English

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

Advertisements
Advertisements

Question

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

Options

  • 0

  • 1

  • π/2

  • π/4

MCQ
Advertisements

Solution

 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
  Is there an error in this question or solution?
Chapter 19: Definite Integrals - MCQ [Page 119]

APPEARS IN

R.D. Sharma Mathematics Volume 1 and 2 [English] Class 12
Chapter 19 Definite Integrals
MCQ | Q 31 | Page 119

RELATED QUESTIONS

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

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

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

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

\[\int\limits_0^{\pi/2} \frac{1}{5 \cos x + 3 \sin x} dx\]

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

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

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

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

\[\int_0^\frac{\pi}{4} \frac{\sin x + \cos x}{3 + \sin2x}dx\]

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

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

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

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

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

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

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

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

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

 


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

Write the coefficient a, b, c of which the value of the integral

\[\int\limits_{- 3}^3 \left( a x^2 + bx + c \right) dx\] is independent.

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


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


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


The value of \[\int\limits_0^{\pi/2} \log\left( \frac{4 + 3 \sin x}{4 + 3 \cos x} \right) dx\] is 

 


Evaluate : \[\int\limits_0^\pi/4 \frac{\sin x + \cos x}{16 + 9 \sin 2x}dx\] .


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


\[\int\limits_1^2 x\sqrt{3x - 2} dx\]


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


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


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


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


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


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


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


Prove that `int_a^b ƒ ("x") d"x" = int_a^bƒ(a + b - "x") d"x" and "hence evaluate" int_(π/6)^(π/3) (d"x")/(1+sqrt(tan "x")`


Evaluate the following using properties of definite integral:

`int_0^1 log (1/x - 1)  "d"x`


Evaluate the following:

Γ(4)


Evaluate \[\int_{0}^{1}x\,dx\].


Evaluate \[\int_{1}^{3}(2x+1)\,dx\].


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×