evaluate-0-pi-sin-2-xdx-

Question Number 48474 by mondodotto@gmail.com last updated on 24/Nov/18

evaluate \int_{0}^{π} \sin^{2} x d x

Answered by hassentimol last updated on 24/Nov/18

Commented by hassentimol last updated on 24/Nov/18

I was in degrees \dots

Answered by tanmay.chaudhury50@gmail.com last updated on 24/Nov/18

\int_{0}^{π} \frac{1 - c o s 2 x}{2} d x

= \frac{1}{2} ∣ x - \frac{s i n 2 x}{2} ∣_{0}^{π}

= \frac{1}{2} \times π

o r \int_{0}^{π} {s i n}^{2} x d x \int_{0}^{2 a} f (x) d x = 2 \int_{0}^{a} f (x) d x w h e n f (x) = f (2 a - x)

= 2 \int_{0}^{\frac{π}{2}} {s i n}^{2} x d x

= 2 \times \frac{1}{2} ∣ x - \frac{s i n 2 x}{2} ∣_{0}^{\frac{π}{2}}

= 2 \times \frac{1}{2} \times \frac{π}{2} = \frac{π}{2}

Commented by mondodotto@gmail.com last updated on 26/Nov/18

sir it is \sin x^{2} not \sin^{2} x

Commented by tanmay.chaudhury50@gmail.com last updated on 26/Nov/18

o k \dots b u t i n q u e s t i o n i t i s {s i n}^{2} x

Commented by hassentimol last updated on 26/Nov/18

Well \dots

\sin {(x)}^{2} = {(\sin (x))}^{2} = \sin^{2} (x) by definition