Advertisements
Advertisements
Question
Prove that:
Advertisements
Solution
\[\int_0^\pi xf\left( \sin x \right)dx = \int_0^\pi \left( \pi - x \right)f\left[ \sin\left( \pi - x \right) \right]dx .................\left[ \int_0^a f\left( x \right)dx = \int_0^a f\left( a - x \right)dx \right]\]
\[ \Rightarrow \int_0^\pi xf\left( \sin x \right)dx = \int_0^\pi \left( \pi - x \right)f\left( \sin x \right)dx\]
\[ \Rightarrow \int_0^\pi xf\left( \sin x \right)dx = \pi \int_0^\pi f\left( \sin x \right)dx - \int_0^\pi xf\left( \sin x \right)dx\]
\[ \Rightarrow 2 \int_0^\pi xf\left( \sin x \right)dx = \pi \int_0^\pi f\left( \sin x \right)dx\]
\[ \Rightarrow \int_0^\pi xf\left( \sin x \right)dx = \frac{\pi}{2} \int_0^\pi f\left( \sin x \right)dx\]
APPEARS IN
RELATED QUESTIONS
Evaluate the following integral:
Write the coefficient a, b, c of which the value of the integral
\[\int_0^\frac{\pi^2}{4} \frac{\sin\sqrt{x}}{\sqrt{x}} dx\] equals
Evaluate : \[\int\limits_0^{2\pi} \cos^5 x dx\] .
\[\int\limits_0^\infty \frac{x}{\left( 1 + x \right)\left( 1 + x^2 \right)} dx\]
\[\int\limits_0^1 \left( \cos^{- 1} x \right)^2 dx\]
\[\int\limits_0^{\pi/2} \left| \sin x - \cos x \right| dx\]
\[\int\limits_0^{2\pi} \cos^7 x dx\]
\[\int\limits_{- \pi/4}^{\pi/4} \left| \tan x \right| dx\]
\[\int\limits_0^{15} \left[ x^2 \right] dx\]
\[\int\limits_0^{\pi/2} \frac{x \sin x \cos x}{\sin^4 x + \cos^4 x} dx\]
\[\int\limits_0^4 x dx\]
\[\int\limits_1^3 \left( 2 x^2 + 5x \right) dx\]
Using second fundamental theorem, evaluate the following:
`int_1^"e" ("d"x)/(x(1 + logx)^3`
Evaluate the following using properties of definite integral:
`int_(-1)^1 log ((2 - x)/(2 + x)) "d"x`
Evaluate `int sqrt((1 + x)/(1 - x)) "d"x`, x ≠1
Find `int x^2/(x^4 + 3x^2 + 2) "d"x`
`int (x + 3)/(x + 4)^2 "e"^x "d"x` = ______.
