What is the integral of sin(1/2)xcos(1/4)x ?

The integral of sin(1/2)xcos(1/4)x is 1/3 sin(1/2)cos(1/4)x^3+C

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integral of sin(1/2)xcos(1/4)x

\int sin (\frac{1}{2}) x cos (\frac{1}{4}) x d x

\frac{1}{3} sin (\frac{1}{2}) cos (\frac{1}{4}) x^{3} + C

Solution steps

\int sin (\frac{1}{2}) x cos (\frac{1}{4}) x d x

Take the constant out:

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

= sin (\frac{1}{2}) cos (\frac{1}{4}) \cdot \int xx d x

Simplify

xx : x^{2}

= sin (\frac{1}{2}) cos (\frac{1}{4}) \cdot \int x^{2} d x

Apply the Power Rule

: \frac{x ^{3}}{3}

= sin (\frac{1}{2}) cos (\frac{1}{4}) \frac{x ^{3}}{3}

Simplify

sin (\frac{1}{2}) cos (\frac{1}{4}) \frac{x ^{3}}{3} : \frac{1}{3} sin (\frac{1}{2}) cos (\frac{1}{4}) x^{3}

= \frac{1}{3} sin (\frac{1}{2}) cos (\frac{1}{4}) x^{3}

Add a constant to the solution

= \frac{1}{3} sin (\frac{1}{2}) cos (\frac{1}{4}) x^{3} + C

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

\frac{\partial}{\partial x} (\frac{x}{x ^{2} + y ^{2}} + 2 x)

\int \frac{4}{x} + \frac{x}{x ^{2} + 5} d x

y = 3 - 9 x^{2}

\frac{d y}{d x}, e^{x} y = 8

What is the integral of sin(1/2)xcos(1/4)x ?
The integral of sin(1/2)xcos(1/4)x is 1/3 sin(1/2)cos(1/4)x^3+C