MCQ
$\int_0^\pi {{{\sin }^2}x\,dx} $ is equal to
- A$\pi $
- ✓$\frac{\pi }{2}$
- C$0$
- DNone of these
✓
Answer
Correct option: B.
$\frac{\pi }{2}$
b
(b) $I = \int_0^\pi {{{\sin }^2}x\,dx = 2\int_0^{\pi /2} {{{\sin }^2}x\,dx} } $,
(b) $I = \int_0^\pi {{{\sin }^2}x\,dx = 2\int_0^{\pi /2} {{{\sin }^2}x\,dx} } $,
$\{\because \,\,\int_{0}^{2a}{f(x)=2\int_{0}^{a}{f(a-x)dx}}$, if $f(2a - x) = f(x) \}$
$I = 2 \times \frac{1}{2} \times \frac{\pi }{2} = \frac{\pi }{2}$.
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