What is the integral from 0 to pi/3 of sin(3x)cos(x) ?

The integral from 0 to pi/3 of sin(3x)cos(x) is 9/16

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integral from 0 to pi/3 of sin(3x)cos(x)

\int_{0}^{\frac{π}{3}} sin (3 x) cos (x) d x

\frac{9}{16}

Solution steps

\int_{0}^{\frac{π}{3}} sin (3 x) cos (x) d x

Use the following identity:

cos (t) sin (s) = \frac{sin ( s + t ) + sin ( s - t )}{2}

= \int_{0}^{\frac{π}{3}} \frac{sin ( 3 x + x ) + sin ( 3 x - x )}{2} d x

Take the constant out:

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

= \frac{1}{2} \cdot \int_{0}^{\frac{π}{3}} sin (3 x + x) + sin (3 x - x) d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{2} (\int_{0}^{\frac{π}{3}} sin (3 x + x) d x + \int_{0}^{\frac{π}{3}} sin (3 x - x) d x)

\int_{0}^{\frac{π}{3}} sin (3 x + x) d x = \frac{3}{8}

\int_{0}^{\frac{π}{3}} sin (3 x - x) d x = \frac{3}{4}

= \frac{1}{2} (\frac{3}{8} + \frac{3}{4})

Simplify

\frac{1}{2} (\frac{3}{8} + \frac{3}{4}) : \frac{9}{16}

= \frac{9}{16}

\int_{0}^{1} 3 t d t

\int_{1}^{3} \frac{1}{x ^{2} + 6 x + 10} d x

\int_{- 1}^{0} x sin (nπ x) d x

\int_{2}^{3} \frac{2}{3 x - 1} d x

\int_{- 5}^{2} x^{2} + 3 x - 10 d x

What is the integral from 0 to pi/3 of sin(3x)cos(x) ?
The integral from 0 to pi/3 of sin(3x)cos(x) is 9/16