What is the integral of 7sin(5x)sin(x) ?

The integral of 7sin(5x)sin(x) is 7/2 (-1/6 sin(6x)+1/4 sin(4x))+C

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integral of 7sin(5x)sin(x)

\int 7 sin (5 x) sin (x) d x

\frac{7}{2} (- \frac{1}{6} sin (6 x) + \frac{1}{4} sin (4 x)) + C

Solution steps

\int 7 sin (5 x) sin (x) d x

Take the constant out:

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

= 7 \cdot \int sin (5 x) sin (x) d x

Rewrite using trig identities

= 7 \cdot \int \frac{1}{2} (- cos (6 x) + cos (4 x)) d x

Take the constant out:

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

= 7 \cdot \frac{1}{2} \cdot \int - cos (6 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

= 7 \cdot \frac{1}{2} (- \int cos (6 x) d x + \int cos (4 x) d x)

\int cos (6 x) d x = \frac{1}{6} sin (6 x)

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

= 7 \cdot \frac{1}{2} (- \frac{1}{6} sin (6 x) + \frac{1}{4} sin (4 x))

Simplify

7 \cdot \frac{1}{2} (- \frac{1}{6} sin (6 x) + \frac{1}{4} sin (4 x)) : \frac{7}{2} (- \frac{1}{6} sin (6 x) + \frac{1}{4} sin (4 x))

= \frac{7}{2} (- \frac{1}{6} sin (6 x) + \frac{1}{4} sin (4 x))

Add a constant to the solution

= \frac{7}{2} (- \frac{1}{6} sin (6 x) + \frac{1}{4} sin (4 x)) + C

x \to 0 lim (4 x - ∣ x ∣)

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

x^{^{''}} - 5 x^{^{'}} + 6 x = cos (t) \cdot e^{- t}

\frac{\partial}{\partial x} (\frac{u ^{2}}{x ^{2}})

What is the integral of 7sin(5x)sin(x) ?
The integral of 7sin(5x)sin(x) is 7/2 (-1/6 sin(6x)+1/4 sin(4x))+C