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

The integral of sin(4x)cos(4x) is -1/8 cos^2(4x)+C

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

\int sin (4 x) cos (4 x) d x

- \frac{1}{8} cos^{2} (4 x) + C

Solution steps

\int sin (4 x) cos (4 x) d x

Apply u-substitution

: \int - \frac{u}{4} d u

= \int - \frac{u}{4} d u

Take the constant out:

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

= - \frac{1}{4} \cdot \int u d u

Apply the Power Rule

: \frac{u ^{2}}{2}

= - \frac{1}{4} \cdot \frac{u ^{2}}{2}

Substitute back

u = cos (4 x)

= - \frac{1}{4} \cdot \frac{cos ^{2} ( 4 x )}{2}

Simplify

- \frac{1}{4} \cdot \frac{cos ^{2} ( 4 x )}{2} : - \frac{1}{8} cos^{2} (4 x)

= - \frac{1}{8} cos^{2} (4 x)

Add a constant to the solution

= - \frac{1}{8} cos^{2} (4 x) + C

\int \frac{3}{2 x ^{2} - 4 x + 10} d x

\frac{d}{d x} (x^{3} 5 - x^{2})

\frac{d y}{d x}, x^{2} + y^{2} = 25

x \to 2 lim ([g (x)]^{3})

What is the integral of sin(4x)cos(4x) ?
The integral of sin(4x)cos(4x) is -1/8 cos^2(4x)+C