What is the integral of sin^2(x)cos(6x) ?

The integral of sin^2(x)cos(6x) is -1/12 (1/4 sin(4x)-1/8 sin(8x))+1/6 sin^2(x)sin(6x)+C

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integral of sin^2(x)cos(6x)

\int sin^{2} (x) cos (6 x) d x

- \frac{1}{12} (\frac{1}{4} sin (4 x) - \frac{1}{8} sin (8 x)) + \frac{1}{6} sin^{2} (x) sin (6 x) + C

Solution steps

\int sin^{2} (x) cos (6 x) d x

Rewrite using trig identities

= \int (1 - cos^{2} (x)) cos (6 x) d x

Apply Integration By Parts

: - \int \frac{1}{6} sin (2 x) sin (6 x) d x + \frac{1}{6} sin^{2} (x) sin (6 x)

= - \int \frac{1}{6} sin (2 x) sin (6 x) d x + \frac{1}{6} sin^{2} (x) sin (6 x)

\int \frac{1}{6} sin (2 x) sin (6 x) d x = \frac{1}{12} (\frac{1}{4} sin (4 x) - \frac{1}{8} sin (8 x))

= - \frac{1}{12} (\frac{1}{4} sin (4 x) - \frac{1}{8} sin (8 x)) + \frac{1}{6} sin^{2} (x) sin (6 x)

Add a constant to the solution

= - \frac{1}{12} (\frac{1}{4} sin (4 x) - \frac{1}{8} sin (8 x)) + \frac{1}{6} sin^{2} (x) sin (6 x) + C

\frac{- 5}{x ^{6}}

\int (1 + x^{2}) (2 x) d x

sec (6 x)

f (x) = \frac{x ^{3} + 8 x ^{2} + x - 7}{x}

n = 1 \sum \infty \frac{3}{( n + 1 ) !}

What is the integral of sin^2(x)cos(6x) ?
The integral of sin^2(x)cos(6x) is -1/12 (1/4 sin(4x)-1/8 sin(8x))+1/6 sin^2(x)sin(6x)+C