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

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

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integral of 7sin(5θ)sin(θ)

\int 7 sin (5 θ) sin (θ) d θ

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

Solution steps

\int 7 sin (5 θ) sin (θ) d θ

Take the constant out:

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

= 7 \cdot \int sin (5 θ) sin (θ) d θ

Rewrite using trig identities

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

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 θ) + cos (4 θ) d θ

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 θ) d θ + \int cos (4 θ) d θ)

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

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

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

Simplify

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

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

Add a constant to the solution

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

a^{2}

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

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

y = x^{2} + 5, (4, 21)

u \to 0 lim (ln (\frac{u + 1}{u}))

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