What is the integral from y to 3y of cos(2x+y) ?

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

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integral from y to 3y of cos(2x+y)

\int_{y}^{3 y} cos (2 x + y) d x

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

Solution steps

\int_{y}^{3 y} cos (2 x + y) d x

Apply u-substitution

: \int_{3 y}^{7 y} cos (u) \frac{1}{2} d u

= \int_{3 y}^{7 y} cos (u) \frac{1}{2} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int_{3 y}^{7 y} cos (u) d u

Use the common integral:

\int cos (u) d u = sin (u)

= \frac{1}{2} [sin (u)]_{3 y}^{7 y}

Compute the boundaries:

sin (7 y) - sin (3 y)

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

\int_{0}^{1} x (1 - x^{4})^{\frac{3}{2}} d x

\int_{1}^{5} x^{4} d x

\int_{1}^{3} 2 x^{2} d x

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

\int_{1}^{3} 3 x d x

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