What is area y=sin(x),y=2sin^2(x),x=0,x= pi/2 ?

The answer to area y=sin(x),y=2sin^2(x),x=0,x= pi/2 is (pi-3sqrt(3))/6+1

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area y=sin(x),y=2sin^2(x),x=0,x= pi/2

area

y = sin (x), y = 2 sin^{2} (x), x = 0, x = \frac{π}{2}

\frac{π - 3 3}{6} + 1

Decimal

0.65757 \dots

Solution steps

Apply the area formula:

\int_{0}^{\frac{π}{2}} sin (x) - 2 sin^{2} (x) d x

Solve

\int_{0}^{\frac{π}{2}} sin (x) - 2 sin^{2} (x) d x : \frac{π - 3 3}{6} + 1

The area is:

= \frac{π - 3 3}{6} + 1

x \to π - lim (\frac{x}{cos ( x ) + 1})

y^{^{''}} + 4 y^{^{'}} + 3 y = 0

x \to 0 - lim (4 + \frac{6}{x ^{2}})

f (x) = x^{2}, at x = - 1

\int \frac{1}{z} d z

What is area y=sin(x),y=2sin^2(x),x=0,x= pi/2 ?
The answer to area y=sin(x),y=2sin^2(x),x=0,x= pi/2 is (pi-3sqrt(3))/6+1