What is the integral of sin^3(4x)cos(4x) ?

The integral of sin^3(4x)cos(4x) is 1/16 sin^4(4x)+C

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integral of sin^3(4x)cos(4x)

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

\frac{1}{16} sin^{4} (4 x) + C

Solution steps

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

Apply u-substitution

Take the constant out:

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

Apply the Power Rule

: \frac{3}{4} u^{\frac{4}{3}}

= \frac{1}{12} \cdot \frac{3}{4} u^{\frac{4}{3}}

Substitute back

u = sin^{3} (4 x)

= \frac{1}{12} \cdot \frac{3}{4} (sin^{3} (4 x))^{\frac{4}{3}}

Simplify

\frac{1}{12} \cdot \frac{3}{4} (sin^{3} (4 x))^{\frac{4}{3}} : \frac{1}{16} sin^{4} (4 x)

= \frac{1}{16} sin^{4} (4 x)

Add a constant to the solution

= \frac{1}{16} sin^{4} (4 x) + C

\int \frac{5 x}{x ^{2} - 1} d x

y^{^{''}} - 8 y^{^{'}} + 16 y = 0, y (0) = 5, y^{^{'}} (0) = 0

h \to 0 lim (\frac{h}{h ^{2} - 7 h + 1})

\frac{d}{d x} (10 0.6 x + 7)

f (x) = x^{2} + 40, at x = 3

What is the integral of sin^3(4x)cos(4x) ?
The integral of sin^3(4x)cos(4x) is 1/16 sin^4(4x)+C