What is the integral from 0 to 2pi of sin(x)sin(x+1) ?

The integral from 0 to 2pi of sin(x)sin(x+1) is picos(1)+(-sin(4pi+1)+sin(1))/4

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integral from 0 to 2pi of sin(x)sin(x+1)

\int_{0}^{2 π} sin (x) sin (x + 1) d x

π cos (1) + \frac{- sin ( 4 π + 1 ) + sin ( 1 )}{4}

Decimal Notation

1.69740 \dots

Solution steps

\int_{0}^{2 π} sin (x) sin (x + 1) d x

Rewrite using trig identities

= \int_{0}^{2 π} \frac{1}{2} (- cos (2 x + 1) + cos (1)) d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int_{0}^{2 π} - cos (2 x + 1) + cos (1) 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

= \frac{1}{2} (- \int_{0}^{2 π} cos (2 x + 1) d x + \int_{0}^{2 π} cos (1) d x)

\int_{0}^{2 π} cos (2 x + 1) d x = \frac{1}{2} (sin (4 π + 1) - sin (1))

\int_{0}^{2 π} cos (1) d x = 2 π cos (1)

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

Simplify

\frac{1}{2} (- \frac{1}{2} (sin (4 π + 1) - sin (1)) + 2 π cos (1)) : π cos (1) + \frac{- sin ( 4 π + 1 ) + sin ( 1 )}{4}

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

\int 2 v d v

f (x) = x^{2} + 55, at x = 3

f (x) = lo g_{4} (x)

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

\int (x^{2} + 4) e^{- x} d x

What is the integral from 0 to 2pi of sin(x)sin(x+1) ?
The integral from 0 to 2pi of sin(x)sin(x+1) is picos(1)+(-sin(4pi+1)+sin(1))/4