What is the integral of sin^2(x)cos(3x) ?

The integral of sin^2(x)cos(3x) is -1/6 (sin(x)-1/5 sin(5x))+1/3 sin^2(x)sin(3x)+C

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integral of sin^2(x)cos(3x)

\int sin^{2} (x) cos (3 x) d x

- \frac{1}{6} (sin (x) - \frac{1}{5} sin (5 x)) + \frac{1}{3} sin^{2} (x) sin (3 x) + C

Solution steps

\int sin^{2} (x) cos (3 x) d x

Rewrite using trig identities

= \int (1 - cos^{2} (x)) cos (3 x) d x

Apply Integration By Parts

: - \int \frac{1}{3} sin (2 x) sin (3 x) d x + \frac{1}{3} sin^{2} (x) sin (3 x)

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

\int \frac{1}{3} sin (2 x) sin (3 x) d x = \frac{1}{6} (sin (x) - \frac{1}{5} sin (5 x))

= - \frac{1}{6} (sin (x) - \frac{1}{5} sin (5 x)) + \frac{1}{3} sin^{2} (x) sin (3 x)

Add a constant to the solution

= - \frac{1}{6} (sin (x) - \frac{1}{5} sin (5 x)) + \frac{1}{3} sin^{2} (x) sin (3 x) + C

\int 2 x (x^{2} + 2)^{8} d x

\int (\frac{4 x + 6}{x ^{2} + 4}) d x

x = 0, x = 5 - y^{2}

\frac{8}{( s + 2 ) ^{2}}

9 y^{^{''}} + 60 y^{^{'}} + 100 y = 0

What is the integral of sin^2(x)cos(3x) ?
The integral of sin^2(x)cos(3x) is -1/6 (sin(x)-1/5 sin(5x))+1/3 sin^2(x)sin(3x)+C