Q.32 Prove that : \(\displaystyle\int_a^b f(x)\,dx = \int_a^b f(a + b - x)\cdot dx\)
Hence, find \(\displaystyle\int_{\pi/6}^{\pi/3}\sin^2 x\cdot dx\)
Best answer
Answer: Property proved; \(\displaystyle\int_{\pi/6}^{\pi/3}\sin^2 x\,dx = \frac{\pi}{12}\).
Proof: In R.H.S. put \(a + b - x = t\), so \(dx = -dt\). When \(x = a,\ t = b\); when \(x = b,\ t = a\).
\(\displaystyle\int_a^b f(a + b - x)\,dx = \int_b^a f(t)(-dt) = \int_a^b f(t)\,dt = \int_a^b f(x)\,dx\) = L.H.S. (a definite integral does not depend on the variable).
Hence part: Let \(I = \displaystyle\int_{\pi/6}^{\pi/3}\sin^2 x\,dx\) ... (1). Here \(a + b = \dfrac{\pi}{6} + \dfrac{\pi}{3} = \dfrac{\pi}{2}\).
By the property, \(I = \displaystyle\int_{\pi/6}^{\pi/3}\sin^2\left(\frac{\pi}{2} - x\right)dx = \int_{\pi/6}^{\pi/3}\cos^2 x\,dx\) ... (2)
Adding (1) and (2): \(2I = \displaystyle\int_{\pi/6}^{\pi/3}(\sin^2 x + \cos^2 x)\,dx = \int_{\pi/6}^{\pi/3}1\,dx = \frac{\pi}{3} - \frac{\pi}{6} = \frac{\pi}{6}\).
\(\therefore I = \dfrac{\pi}{12}\).
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