Question:

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

Updated On: Oct 7, 2023
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Solution and Explanation

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

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

\(2I\) = \(∫_0^{\frac \pi2}1\ dx\)

\(2I\) = \([x]_0^{\frac \pi2}\)

\(2I\) = \(\frac \pi2\)

\(I\) = \(\frac \pi4\)

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