Evaluate the definite integral \(\int_0^1 sin^2x \,dx\).

(a) –\(\frac{π}{2}\)

(b) π

(c) \(\frac{π}{4}\)

(d) \(\frac{π}{6}\)

This question was addressed to me by my school principal while I was bunking the class.

Asked question is from Fundamental Theorem of Calculus-2 topic in chapter Integrals of Mathematics – Class 12

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The correct answer is (c) \(\frac{π}{4}\)

To explain I would say: Let \(I=\int_0^{π/2}sin^{2}x \,dx\)

F(x)=\(\int sin^2x \,dx\)

=\(\int \frac{(1-cos2x)}{2} \,dx\)

=\(\frac{1}{2} (x-\frac{sin2x}{2})\)

Applying the limits, we get

\(I=F(\frac{π}{2})-F(0)=\frac{1}{2} (\frac{π}{2}-\frac{sinπ}{2})-\frac{1}{2} (0-\frac{sin0}{2})\)

=\(\frac{1}{4} (π-0)-0=\frac{π}{4}\).