English

Π / 2 ∫ − π / 2 X Cos 2 X D X . - Mathematics

Advertisements
Advertisements

Question

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

 

Sum
Advertisements

Solution

\[\text{We have}, \]

\[I = \int_{- \frac{\pi}{2}}^\frac{\pi}{2} x \cos^2 x\ d x\]

\[Let f\left( x \right) = x \cos^2 x\]

\[ \Rightarrow f\left( - x \right) = \left( - x \right) \cos^2 \left( - x \right)\]

\[ = - x \cos^2 x\]

\[ \therefore f\left( - x \right) = - f\left( x \right)\]

\[i . e . , f\left( x \right) \text{is odd function}\]

\[\text{We know that} \int_{- a}^a f\left( x \right) d x = 0 , \text{if }f\left( x \right) \text{is odd function} . \]

\[ \therefore I = \int_{- \frac{\pi}{2}}^\frac{\pi}{2} x \cos^2 x\ d x = 0\]

shaalaa.com
Definite Integrals
  Is there an error in this question or solution?
Chapter 20: Definite Integrals - Very Short Answers [Page 115]

APPEARS IN

RD Sharma Mathematics [English] Class 12
Chapter 20 Definite Integrals
Very Short Answers | Q 6 | Page 115

RELATED QUESTIONS

\[\int\limits_4^9 \frac{1}{\sqrt{x}} dx\]

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

\[\int\limits_{- \pi/4}^{\pi/4} \frac{1}{1 + \sin x} dx\]

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

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

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

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

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

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

\[\int_0^\frac{\pi}{4} \frac{\sin^2 x \cos^2 x}{\left( \sin^3 x + \cos^3 x \right)^2}dx\]

\[\int\limits_0^9 f\left( x \right) dx, where f\left( x \right) \begin{cases}\sin x & , & 0 \leq x \leq \pi/2 \\ 1 & , & \pi/2 \leq x \leq 3 \\ e^{x - 3} & , & 3 \leq x \leq 9\end{cases}\]

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

If f is an integrable function, show that

\[\int\limits_{- a}^a x f\left( x^2 \right) dx = 0\]

 


\[\int\limits_0^1 \left( 3 x^2 + 5x \right) dx\]

\[\int\limits_1^4 \left( 3 x^2 + 2x \right) dx\]

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

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

\[\int\limits_0^{\pi/2} \sqrt{1 - \cos 2x}\ dx .\]

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

 

 


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

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

The value of \[\int\limits_0^\pi \frac{x \tan x}{\sec x + \cos x} dx\] is __________ .


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


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

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

 


\[\int\limits_{- 1}^1 \left| 1 - x \right| dx\]  is equal to

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

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

 


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


\[\int\limits_{- a}^a \frac{x e^{x^2}}{1 + x^2} dx\]


\[\int\limits_0^{\pi/2} \frac{1}{1 + \cot^7 x} dx\]


\[\int\limits_0^\pi x \sin x \cos^4 x dx\]


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


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


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


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


\[\int\limits_0^4 x dx\]


Evaluate the following using properties of definite integral:

`int_(- pi/4)^(pi/4) x^3 cos^3 x  "d"x`


Evaluate the following:

`int_0^oo "e"^(-4x) x^4  "d"x`


Choose the correct alternative:

`int_0^1 (2x + 1)  "d"x` is


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×