Question

By using properties of definite integrals, evaluate the integrals
Let ${\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}si{n}^{2}xdx.$

Answer 1

Let ${\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}si{n}^{2}xdx.$
Here,$f\left(x\right)=si{n}^{2}xf(-x)=si{n}^2(-x)=\left[sin\right(-x){]}^2=(-sinx{)}^2=si{n}^{2}x=f\left(x\right)\therefore \text{f(x) is an even function}.I={\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}si{n}^{2}xdx=2{\int }_0^{\frac{\pi }{2}}si{n}^{2}xdx[\because {\int }_{-a}^{a}f(x)dx={\int }_0^{a}f(x)dx,iff(x\left)iseven\right]=2{\int }_0^{\frac{\pi }{2}}\left[\frac{1-cos2x}{2}\right]dx(\because cos2x=1-2si{n}^{2}x)=f_0^{\frac{\pi }{2}}(1-cos2x)dx={[x-\frac{sin2x}{2}]}_0^{\frac{\pi }{2}}=[\frac{\pi }{2}-\frac{sin\pi }{2}]-(0-0)=\frac{\pi }{2}-0=\frac{\pi }{2}$