What is the integral from 1 to 2 of (1+cos(x))^2 ?

The integral from 1 to 2 of (1+cos(x))^2 is -2sin(1)+3/2+(sin(4)+7sin(2))/4

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integral from 1 to 2 of (1+cos(x))^2

\int_{1}^{2} (1 + cos (x))^{2} d x

- 2 sin (1) + \frac{3}{2} + \frac{sin ( 4 ) + 7 sin ( 2 )}{4}

Decimal

1.21912 \dots

Solution steps

\int_{1}^{2} (1 + cos (x))^{2} d x

Expand

(1 + cos (x))^{2} : 1 + 2 cos (x) + cos^{2} (x)

= \int_{1}^{2} 1 + 2 cos (x) + cos^{2} (x) d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{1}^{2} 1 d x + \int_{1}^{2} 2 cos (x) d x + \int_{1}^{2} cos^{2} (x) d x

\int_{1}^{2} 1 d x = 1

\int_{1}^{2} 2 cos (x) d x = 2 (sin (2) - sin (1))

\int_{1}^{2} cos^{2} (x) d x = \frac{1}{2} (1 + \frac{1}{2} (sin (4) - sin (2)))

= 1 + 2 (sin (2) - sin (1)) + \frac{1}{2} (1 + \frac{1}{2} (sin (4) - sin (2)))

Simplify

1 + 2 (sin (2) - sin (1)) + \frac{1}{2} (1 + \frac{1}{2} (sin (4) - sin (2))) : - 2 sin (1) + \frac{3}{2} + \frac{sin ( 4 ) + 7 sin ( 2 )}{4}

= - 2 sin (1) + \frac{3}{2} + \frac{sin ( 4 ) + 7 sin ( 2 )}{4}

What is the integral from 1 to 2 of (1+cos(x))^2 ?
The integral from 1 to 2 of (1+cos(x))^2 is -2sin(1)+3/2+(sin(4)+7sin(2))/4