What is the integral of 64e^{4θ}sin(4θ) ?

The integral of 64e^{4θ}sin(4θ) is 64((e^{4θ}sin(4θ))/8-(e^{4θ}cos(4θ))/8)+C

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integral of 64e^{4θ}sin(4θ)

\int 64 e^{4 θ} sin (4 θ) d θ

64 (\frac{e ^{4 θ} sin ( 4 θ )}{8} - \frac{e ^{4 θ} cos ( 4 θ )}{8}) + C

Solution steps

\int 64 e^{4 θ} sin (4 θ) d θ

Take the constant out:

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

= 64 \cdot \int e^{4 θ} sin (4 θ) d θ

Apply Integration By Parts

= 64 (- \frac{1}{4} e^{4 θ} cos (4 θ) - \int - e^{4 θ} cos (4 θ) d θ)

Take the constant out:

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

= 64 (- \frac{1}{4} e^{4 θ} cos (4 θ) - (- \int e^{4 θ} cos (4 θ) d θ))

Apply Integration By Parts

= 64 (- \frac{1}{4} e^{4 θ} cos (4 θ) - (- (\frac{1}{4} e^{4 θ} sin (4 θ) - \int e^{4 θ} sin (4 θ) d θ)))

Therefore

64 \cdot \int e^{4 θ} sin (4 θ) d θ = 64 (- \frac{1}{4} e^{4 θ} cos (4 θ) - (- (\frac{1}{4} e^{4 θ} sin (4 θ) - \int e^{4 θ} sin (4 θ) d θ)))

Isolate

\int e^{4 θ} sin (4 θ) d θ

= 64 (\frac{e ^{4 θ} sin ( 4 θ )}{8} - \frac{e ^{4 θ} cos ( 4 θ )}{8})

Add a constant to the solution

= 64 (\frac{e ^{4 θ} sin ( 4 θ )}{8} - \frac{e ^{4 θ} cos ( 4 θ )}{8}) + C

What is the integral of 64e^{4θ}sin(4θ) ?
The integral of 64e^{4θ}sin(4θ) is 64((e^{4θ}sin(4θ))/8-(e^{4θ}cos(4θ))/8)+C