If ∫0π xf(sin⁡x)dx=A∫0π/2 f(sin⁡x)dx ,then A is

A
$2 π$
B
$π$
C
$π / 4$
D
0

Solution:

Let $I = \int_{0}^{π} x f (\sin x) d x .$

$\begin{array}{l} I = \int_{0}^{π} (π - x) f (\sin (π - x)) d x [∵ \int_{0}^{a} f (x) d x = \int_{0}^{a} f (a - x) d x] \\ \Rightarrow I = π \int_{0}^{π} f (\sin x) d x - I \\ \Rightarrow I = \frac{π}{2} \int_{0}^{π} f (\sin x) d x \end{array}$

$\begin{array}{l} \Rightarrow I = \frac{π}{2} \times 2 \int_{0}^{π / 2} f (\sin x) d x [∵ \int_{0}^{2 a} f (x) d x = 2 \int_{0}^{a} f (x) d x if, f (2 a - x) = f (x)] \\ \Rightarrow I = π \int_{0}^{π / 2} f (\sin x) \\ ∴ \int_{0}^{π} x f (\sin x) d x = A \int_{0}^{π / 2} f (\sin x) d x \\ \Rightarrow π \int_{0}^{π / 2} f (\sin x) d x = A \int_{0}^{π / 2} f (\sin x) d x \\ = A = π \end{array}$

If $\int_{0}^{π} x f (\sin x) d x = A \int_{0}^{π / 2} f (\sin x) d x$ ,then $A$ is

Solution:

Related content