What is the integral of 5(5-4cos(t))^{1/4}sin(t) ?

The integral of 5(5-4cos(t))^{1/4}sin(t) is (5-4cos(t))^{5/4}+C

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integral of 5(5-4cos(t))^{1/4}sin(t)

\int 5 (5 - 4 cos (t))^{\frac{1}{4}} sin (t) d t

(5 - 4 cos (t))^{\frac{5}{4}} + C

Solution steps

\int 5 (5 - 4 cos (t))^{\frac{1}{4}} sin (t) d t

Take the constant out:

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

= 5 \cdot \int (5 - 4 cos (t))^{\frac{1}{4}} sin (t) d t

Apply u-substitution

: \int \frac{u ^{\frac{1}{4}}}{4} d u

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

Apply the Power Rule

: \frac{4}{5} u^{\frac{5}{4}}

= 5 \cdot \frac{1}{4} \cdot \frac{4}{5} u^{\frac{5}{4}}

Substitute back

u = 5 - 4 cos (t)

= 5 \cdot \frac{1}{4} \cdot \frac{4}{5} (5 - 4 cos (t))^{\frac{5}{4}}

Simplify

5 \cdot \frac{1}{4} \cdot \frac{4}{5} (5 - 4 cos (t))^{\frac{5}{4}} : (5 - 4 cos (t))^{\frac{5}{4}}

= (5 - 4 cos (t))^{\frac{5}{4}}

Add a constant to the solution

= (5 - 4 cos (t))^{\frac{5}{4}} + C

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What is the integral of 5(5-4cos(t))^{1/4}sin(t) ?
The integral of 5(5-4cos(t))^{1/4}sin(t) is (5-4cos(t))^{5/4}+C