If $\int_0^a f(x)dx = k$, then the value of $\int_0^a f(kx)dx$ is equal to
Step-by-Step Solution
Key Concept: Use trigonometric identities or the reduction formula $\sin^2 x = \frac{1 - \cos 2x}{2}$ for definite integration over a period
Using the identity $\sin^3 x = \frac{3}{8}\sin x - \frac{1}{8}\sin 3x$, we compute $\int_0^{\pi} \sin^2 x\, dx = \frac{3}{8}\int_0^{\pi} \sin x\, dx - \frac{1}{8}\int_0^{\pi} \sin 3x\, dx$. The first integral evaluates to $[-\cos x]_0^{\pi} = 2$, and the second to $[-\frac{1}{3}\cos 3x]_0^{\pi} = 0$. Therefore $\int_0^{\pi} \sin^2 x\, dx = \frac{3}{8}(2) = \frac{3}{4}$. However, using the direct formula $\int_0^{\pi} \sin^2 x\, dx = [\frac{x}{2} - \frac{\sin 2x}{4}]_0^{\pi} = \frac{\pi}{2}$.
Correct Answer: Shown to equal $\frac{\pi}{2}$