What is the integral of sin(8x)cos(8x) ?

The integral of sin(8x)cos(8x) is -1/16 cos^2(8x)+C

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integral of sin(8x)cos(8x)

\int sin (8 x) cos (8 x) d x

- \frac{1}{16} cos^{2} (8 x) + C

Solution steps

\int sin (8 x) cos (8 x) d x

Apply u-substitution

: \int - \frac{u}{8} d u

= \int - \frac{u}{8} d u

Take the constant out:

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

= - \frac{1}{8} \cdot \int u d u

Apply the Power Rule

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

= - \frac{1}{8} \cdot \frac{u ^{2}}{2}

Substitute back

u = cos (8 x)

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

Simplify

- \frac{1}{8} \cdot \frac{cos ^{2} ( 8 x )}{2} : - \frac{1}{16} cos^{2} (8 x)

= - \frac{1}{16} cos^{2} (8 x)

Add a constant to the solution

= - \frac{1}{16} cos^{2} (8 x) + C

(x - 1)^{2}

\frac{d}{d x} (sec^{2} (x) d x)

(6 - 10 y + e^{- 5 x}) d x - 2 d y = 0

\frac{2 - s}{s ^{2} + 6 s + 34}

\frac{1 - x}{1 + x}

What is the integral of sin(8x)cos(8x) ?
The integral of sin(8x)cos(8x) is -1/16 cos^2(8x)+C