What is the integral of 16sin(x)cos^2(x) ?

The integral of 16sin(x)cos^2(x) is -16/3 cos^3(x)+C

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integral of 16sin(x)cos^2(x)

\int 16 sin (x) cos^{2} (x) d x

- \frac{16}{3} cos^{3} (x) + C

Solution steps

\int 16 sin (x) cos^{2} (x) d x

Take the constant out:

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

= 16 \cdot \int sin (x) cos^{2} (x) d x

Apply u-substitution

: \int - u^{2} d u

= 16 \cdot \int - u^{2} d u

Take the constant out:

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

= 16 (- \int u^{2} d u)

Apply the Power Rule

: \frac{u ^{3}}{3}

= 16 (- \frac{u ^{3}}{3})

Substitute back

u = cos (x)

= 16 (- \frac{cos ^{3} ( x )}{3})

Simplify

16 (- \frac{cos ^{3} ( x )}{3}) : - \frac{16}{3} cos^{3} (x)

= - \frac{16}{3} cos^{3} (x)

Add a constant to the solution

= - \frac{16}{3} cos^{3} (x) + C

x \to - 3 lim (x^{2})

f (x) = 1 + 4 - x^{2}, at x = 0

\int e^{4 x} sin (3 x) d x

x \to 0 + lim (\frac{7}{x} - 7 ∣ x ∣)

\frac{1}{x - 1}, 12

What is the integral of 16sin(x)cos^2(x) ?
The integral of 16sin(x)cos^2(x) is -16/3 cos^3(x)+C