What is the integral of-15cos^3(t)sin(t) ?

The integral of-15cos^3(t)sin(t) is 15/4 cos^4(t)+C

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integral of-15cos^3(t)sin(t)

\int - 15 cos^{3} (t) sin (t) d t

\frac{15}{4} cos^{4} (t) + C

Solution steps

\int - 15 cos^{3} (t) sin (t) d t

Take the constant out:

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

= - 15 \cdot \int cos^{3} (t) sin (t) d t

Apply u-substitution

: \int - u^{3} d u

= - 15 \cdot \int - u^{3} d u

Take the constant out:

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

= - 15 (- \int u^{3} d u)

Apply the Power Rule

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

= - 15 (- \frac{u ^{4}}{4})

Substitute back

u = cos (t)

= - 15 (- \frac{cos ^{4} ( t )}{4})

Simplify

- 15 (- \frac{cos ^{4} ( t )}{4}) : \frac{15}{4} cos^{4} (t)

= \frac{15}{4} cos^{4} (t)

Add a constant to the solution

= \frac{15}{4} cos^{4} (t) + C

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What is the integral of-15cos^3(t)sin(t) ?
The integral of-15cos^3(t)sin(t) is 15/4 cos^4(t)+C