What is the integral of cos(7x)cos(3x) ?

The integral of cos(7x)cos(3x) is 1/2 (1/10 sin(10x)+1/4 sin(4x))+C

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integral of cos(7x)cos(3x)

\int cos (7 x) cos (3 x) d x

\frac{1}{2} (\frac{1}{10} sin (10 x) + \frac{1}{4} sin (4 x)) + C

Solution steps

\int cos (7 x) cos (3 x) d x

Rewrite using trig identities

= \int \frac{1}{2} (cos (10 x) + cos (4 x)) 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 cos (10 x) + cos (4 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 cos (10 x) d x + \int cos (4 x) d x)

\int cos (10 x) d x = \frac{1}{10} sin (10 x)

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

= \frac{1}{2} (\frac{1}{10} sin (10 x) + \frac{1}{4} sin (4 x))

Add a constant to the solution

= \frac{1}{2} (\frac{1}{10} sin (10 x) + \frac{1}{4} sin (4 x)) + C

\int e^{- s x} x^{2} d x

x^{2} \cdot ln (7 x)

\int_{\frac{1}{5}}^{7} 10 x ln (5 x) d x

f (t) = sin (5 t)

\int 125 x^{2} cos (5 x) d x

What is the integral of cos(7x)cos(3x) ?
The integral of cos(7x)cos(3x) is 1/2 (1/10 sin(10x)+1/4 sin(4x))+C