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

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

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

\[\int\limits_0^{\pi/6} \cos x \cos 2x\ dx\]

\[\int\limits_0^{\pi/2} \cos^4\ x\ dx\]

 


\[\int\limits_1^2 \frac{x + 3}{x \left( x + 2 \right)} dx\]

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

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

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

\[\int\limits_2^4 \frac{x}{x^2 + 1} dx\]

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

\[\int\limits_0^1 \frac{24 x^3}{\left( 1 + x^2 \right)^4} dx\]

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

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

 


\[\int_{- \frac{\pi}{2}}^\frac{\pi}{2} \left( 2\sin\left| x \right| + \cos\left| x \right| \right)dx\]

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

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

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

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

\[\int\limits_{- \pi/2}^{\pi/2} \log\left( \frac{2 - \sin x}{2 + \sin x} \right) dx\]

Evaluate the following integral:

\[\int_{- 1}^1 \left| xcos\pi x \right|dx\]

 


\[\int\limits_0^2 e^x dx\]

\[\int\limits_0^{\pi/4} \tan^2 x\ dx .\]

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

Evaluate each of the following  integral:

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

 


Evaluate each of the following integral:

\[\int_e^{e^2} \frac{1}{x\log x}dx\]

Evaluate each of the following integral:

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

 


Solve each of the following integral:

\[\int_2^4 \frac{x}{x^2 + 1}dx\]

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


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

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

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


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


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


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


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


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")`


Choose the correct alternative:

If n > 0, then Γ(n) is


Evaluate `int (3"a"x)/("b"^2 + "c"^2x^2) "d"x`


Evaluate `int (x^2"d"x)/(x^4 + x^2 - 2)`


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×