Question

By using the properties of definite integrals, evaluate the integral: $∫_0^{\frac \pi2}cos^2x\ dx$

Answer 1

I=$$∫_0^{\frac \pi2}cos^2x\ dx ...(1)

⇒ $I$ =$∫_0^{\frac \pi2}cos^2(\frac \pi2-x)\ dx$ $(∫_0^aƒ(x)dx = ƒ(a-x)dx)$

⇒$I$ = $∫_0^{\frac \pi2}sin^2x\ dx$ ...(2)

Adding (1) and (2), we obtain

$2I$ =$∫_0^{\frac \pi2}(sin ^2x+cos^2x)\ dx$

⇒$2I$ = $∫_0^{\frac \pi2}1\ dx$

⇒$2I$ = $[x]_0^{\frac \pi2}$

⇒$2I$ = $\frac \pi2$

⇒$I$ = $\frac \pi4$