Question

By using the properties of definite integrals, evaluate the integral: \(∫^\frac{π}{2}_\frac{-π}{2}sin^2xdx\)

Answer 1

Let \(∫^\frac{π}{2}_\frac{-π}{2}sin^2xdx\)

\(As\, sin^2(−x)=(sin(−x))^2=(−sinx)^2=sin^2x\,therefore,\,sin^2x\,\, is\,\, an\,\, even\,\, function.\)

\(It\,\, is\,\, known\,\, that\,\, if\,\, f(x)\,is\,\, an\,\, even\, function,then\,\, ∫^a_{-a}ƒ(x)dx=2∫^a_0ƒ(x)dx\)

\(I=2∫_0^{π}{2} sin^2xdx\)

\(=2∫_0^{π}{2} \frac{1-cos2x}{2}dx\)

\(=∫_0^\frac{π}{2}(1-cos2x)dx\)

\(=[x-\frac{sin2x}{2}]^\frac{π}{2}_0\)

\(=\frac{π}{2}\)