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

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

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

\int_{0}^{2 π} ∣ sin (x + 1) ∣ d x

4 + cos (1) - cos (2 π + 1)

Decimal

4

Solution steps

\int_{0}^{2 π} ∣ sin (x + 1) ∣ d x

Eliminate Absolutes

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

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

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

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

= 1 + cos (1) + 2 - cos (2 π + 1) + 1

Simplify

= 4 + cos (1) - cos (2 π + 1)

\int_{0}^{\frac{π}{2}} sin (4 x) d x

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

\int_{0}^{1} \frac{x}{1 - x} d x

\int_{e}^{e^{2}} \frac{ln ( x )}{x} d x

\int_{0}^{1} \frac{x ^{2}}{1 + x ^{3}} d x

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