What is the integral of cos(2x)sin(x) ?

The integral of cos(2x)sin(x) is 1/2 (-1/3 cos(3x)+cos(x))+C

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integral of cos(2x)sin(x)

\int cos (2 x) sin (x) d x

\frac{1}{2} (- \frac{1}{3} cos (3 x) + cos (x)) + C

Solution steps

\int cos (2 x) sin (x) d x

Use the following identity:

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

= \int \frac{sin ( x + 2 x ) + sin ( x - 2 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 sin (x + 2 x) + sin (x - 2 x) d x

Simplify

sin (x + 2 x) + sin (x - 2 x) : sin (3 x) - sin (x)

= \frac{1}{2} \cdot \int sin (3 x) - sin (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 sin (3 x) d x - \int sin (x) d x)

\int sin (3 x) d x = - \frac{1}{3} cos (3 x)

\int sin (x) d x = - cos (x)

= \frac{1}{2} (- \frac{1}{3} cos (3 x) - (- cos (x)))

Simplify

= \frac{1}{2} (- \frac{1}{3} cos (3 x) + cos (x))

Add a constant to the solution

= \frac{1}{2} (- \frac{1}{3} cos (3 x) + cos (x)) + C

\int (∣ x ∣ + 1) d x

\frac{x ^{2}}{1 + x}, 0

8 (sin (t) y^{^{'}} + cos (t) y) = cos (t) sin^{5} (t)

\frac{1}{1 + 5 x ^{2}}

f (x) = 5 x + 1, at x = 7

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