The value of the integral ∫ limits0π x ℓ n sin x d x is

Question

The value of the integral ∫ limits0π x ℓ n sin x d x is (A) -(π2/2) ℓ ln 2 (B) (π2/2) ln 2 (C) π2 ℓ n 2 (D) -π2 ℓ n 2

Tardigrade Admin · Accepted Answer

I=∫ limits0π x ℓ n sin x d x I=∫ limits0π(π-x) ℓ n sin (π-x) d x =∫ limits0π(π-x) ℓ n sin x d x Add 2 I=π ∫ limits0π ℓ n sin x d x I=(π/2) ∫ limits0π ℓ n sin x d x =-(π2/2) ln 2

Q. The value of the integral $0 \int π x ℓ n sin x d x$ is

$- \frac{π ^{2}}{2} ℓ ln 2$

$\frac{π ^{2}}{2} ln 2$

$π^{2} ℓ n 2$

$- π^{2} ℓ n 2$

$I = 0 \int π x ℓ n sin x d x$
$I = 0 \int π (π - x) ℓ n sin (π - x) d x$
$= 0 \int π (π - x) ℓ n sin x d x$
Add $2 I = π 0 \int π ℓ n sin x d x$
$I = \frac{π}{2} 0 \int π ℓ n sin x d x$
$= - \frac{π ^{2}}{2} ln 2$

Q. The value of the integral 0∫π​xℓnsinxdx is

−2π2​ℓln2

2π2​ln2

π2ℓn2

−π2ℓn2