What is the integral from 0 to pi of 5(5-4cos(x))^{1/4}sin(x) ?

The integral from 0 to pi of 5(5-4cos(x))^{1/4}sin(x) is 9*3^{1/2}-1

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integral from 0 to pi of 5(5-4cos(x))^{1/4}sin(x)

\int_{0}^{π} 5 (5 - 4 cos (x))^{\frac{1}{4}} sin (x) d x

9 \cdot 3^{\frac{1}{2}} - 1

Decimal

14.58845 \dots

Solution steps

\int_{0}^{π} 5 (5 - 4 cos (x))^{\frac{1}{4}} sin (x) d x

Take the constant out:

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

= 5 \cdot \int_{0}^{π} (5 - 4 cos (x))^{\frac{1}{4}} sin (x) d x

Apply u-substitution

: \int_{1}^{9} \frac{u ^{\frac{1}{4}}}{4} d u

= 5 \cdot \int_{1}^{9} \frac{u ^{\frac{1}{4}}}{4} d u

Take the constant out:

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

= 5 \cdot \frac{1}{4} \cdot \int_{1}^{9} u^{\frac{1}{4}} d u

Apply the Power Rule

: [\frac{4}{5} u^{\frac{5}{4}}]_{1}^{9}

= 5 \cdot \frac{1}{4} [\frac{4}{5} u^{\frac{5}{4}}]_{1}^{9}

Simplify

5 \cdot \frac{1}{4} [\frac{4}{5} u^{\frac{5}{4}}]_{1}^{9} : \frac{5}{4} [\frac{4}{5} u^{\frac{5}{4}}]_{1}^{9}

= \frac{5}{4} [\frac{4}{5} u^{\frac{5}{4}}]_{1}^{9}

Compute the boundaries:

\frac{36 \cdot 3 ^{\frac{1}{2}}}{5} - \frac{4}{5}

= \frac{5}{4} (\frac{36 \cdot 3 ^{\frac{1}{2}}}{5} - \frac{4}{5})

Simplify

= 9 \cdot 3^{\frac{1}{2}} - 1

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What is the integral from 0 to pi of 5(5-4cos(x))^{1/4}sin(x) ?
The integral from 0 to pi of 5(5-4cos(x))^{1/4}sin(x) is 9*3^{1/2}-1