English

Prove That: ∫ π 0 X F ( Sin X ) D X = π 2 ∫ π 0 F ( Sin X ) D X - Mathematics

Advertisements
Advertisements

Question

Prove that:

\[\int_0^\pi xf\left( \sin x \right)dx = \frac{\pi}{2} \int_0^\pi f\left( \sin x \right)dx\]
Sum
Advertisements

Solution

\[\int_0^\pi xf\left( \sin x \right)dx = \int_0^\pi \left( \pi - x \right)f\left[ \sin\left( \pi - x \right) \right]dx .................\left[ \int_0^a f\left( x \right)dx = \int_0^a f\left( a - x \right)dx \right]\]
\[ \Rightarrow \int_0^\pi xf\left( \sin x \right)dx = \int_0^\pi \left( \pi - x \right)f\left( \sin x \right)dx\]
\[ \Rightarrow \int_0^\pi xf\left( \sin x \right)dx = \pi \int_0^\pi f\left( \sin x \right)dx - \int_0^\pi xf\left( \sin x \right)dx\]
\[ \Rightarrow 2 \int_0^\pi xf\left( \sin x \right)dx = \pi \int_0^\pi f\left( \sin x \right)dx\]
\[ \Rightarrow \int_0^\pi xf\left( \sin x \right)dx = \frac{\pi}{2} \int_0^\pi f\left( \sin x \right)dx\]

shaalaa.com
Definite Integrals
  Is there an error in this question or solution?
Chapter 20: Definite Integrals - Exercise 20.5 [Page 96]

APPEARS IN

RD Sharma Mathematics [English] Class 12
Chapter 20 Definite Integrals
Exercise 20.5 | Q 49 | Page 96

RELATED QUESTIONS

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

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

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

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

\[\int\limits_0^1 x \left( 1 - x \right)^5 dx\]

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

\[\int_\frac{\pi}{6}^\frac{\pi}{3} \left( \tan x + \cot x \right)^2 dx\]

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

\[\int\limits_4^{12} x \left( x - 4 \right)^{1/3} dx\]

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

\[\int\limits_1^4 f\left( x \right) dx, where f\left( x \right) = \begin{cases}7x + 3 & , & \text{if }1 \leq x \leq 3 \\ 8x & , & \text{if }3 \leq x \leq 4\end{cases}\]


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

 


\[\int\limits_0^\pi \frac{x}{1 + \cos \alpha \sin x} dx, 0 < \alpha < \pi\]

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

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

 


If f (x) is a continuous function defined on [0, 2a]. Then, prove that

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

 


\[\int\limits_2^3 x^2 dx\]

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

\[\int\limits_0^3 \frac{1}{x^2 + 9} dx .\]

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

Evaluate each of the following  integral:

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

 


Evaluate each of the following integral:

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

 


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

\[\int\limits_0^{\pi/2} \frac{\cos x}{\left( 2 + \sin x \right)\left( 1 + \sin x \right)} dx\] equals

`int_0^1 sqrt((1 - "x")/(1 + "x")) "dx"`


The derivative of \[f\left( x \right) = \int\limits_{x^2}^{x^3} \frac{1}{\log_e t} dt, \left( x > 0 \right),\] is

 


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


\[\int\limits_1^2 \frac{1}{x^2} e^{- 1/x} dx\]


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


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


Find : `∫_a^b logx/x` 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")`


Using second fundamental theorem, evaluate the following:

`int_1^2 (x - 1)/x^2  "d"x`


Evaluate the following using properties of definite integral:

`int_(-1)^1 log ((2 - x)/(2 + x))  "d"x`


Evaluate the following:

Γ(4)


Evaluate the following integrals as the limit of the sum:

`int_0^1 (x + 4)  "d"x`


Evaluate the following integrals as the limit of the sum:

`int_1^3 (2x + 3)  "d"x`


Choose the correct alternative:

`int_0^1 (2x + 1)  "d"x` is


If x = `int_0^y "dt"/sqrt(1 + 9"t"^2)` and `("d"^2y)/("d"x^2)` = ay, then a equal to ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×