What is the integral from 0 to pi/4 of 18cos(2t) ?

The integral from 0 to pi/4 of 18cos(2t) is 9

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integral from 0 to pi/4 of 18cos(2t)

\int_{0}^{\frac{π}{4}} 18 cos (2 t) d t

9

Solution steps

\int_{0}^{\frac{π}{4}} 18 cos (2 t) d t

Take the constant out:

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

= 18 \cdot \int_{0}^{\frac{π}{4}} cos (2 t) d t

Apply u-substitution

: \int_{0}^{\frac{π}{2}} cos (u) \frac{1}{2} d u

= 18 \cdot \int_{0}^{\frac{π}{2}} 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

= 18 \cdot \frac{1}{2} \cdot \int_{0}^{\frac{π}{2}} cos (u) d u

Use the common integral:

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

= 18 \cdot \frac{1}{2} [sin (u)]_{0}^{\frac{π}{2}}

Simplify

18 \cdot \frac{1}{2} [sin (u)]_{0}^{\frac{π}{2}} : 9 [sin (u)]_{0}^{\frac{π}{2}}

= 9 [sin (u)]_{0}^{\frac{π}{2}}

Compute the boundaries:

1

= 9 \cdot 1

Simplify

= 9

\int_{5}^{\infty} x e^{- 5 x} d x

\int_{0}^{\frac{1}{2}} e^{- x} d x

\int_{1}^{2} \frac{1}{x ^{2} + 2 x + 5} d x

\int_{0}^{10} 2 e^{- 2 x} d x

\int_{0}^{2} 20 t^{2} e^{- t} d t

What is the integral from 0 to pi/4 of 18cos(2t) ?
The integral from 0 to pi/4 of 18cos(2t) is 9