What is the integral from 0 to 1 of 9cos((pit)/2) ?

The integral from 0 to 1 of 9cos((pit)/2) is (18)/pi

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integral from 0 to 1 of 9cos((pit)/2)

\int_{0}^{1} 9 cos (\frac{π t}{2}) d t

\frac{18}{π}

Solution steps

\int_{0}^{1} 9 cos (\frac{π t}{2}) d t

Take the constant out:

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

= 9 \cdot \int_{0}^{1} cos (\frac{π t}{2}) d t

Apply u-substitution

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

= 9 \cdot \int_{0}^{\frac{π}{2}} cos (u) \frac{2}{π} d u

Take the constant out:

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

= 9 \cdot \frac{2}{π} \cdot \int_{0}^{\frac{π}{2}} cos (u) d u

Use the common integral:

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

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

Simplify

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

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

Compute the boundaries:

1

= \frac{18}{π} \cdot 1

Simplify

= \frac{18}{π}

\int \frac{1}{x ^{2} + 36} d x

ln (3)

\frac{d}{d x} (ln (\frac{1}{1 - x}))

y = 4 x^{2}, y = x^{2} + 6

x \to 7 lim (x + 2)

What is the integral from 0 to 1 of 9cos((pit)/2) ?
The integral from 0 to 1 of 9cos((pit)/2) is (18)/pi