What is the integral of e^{4x}sin(bx) ?

The integral of e^{4x}sin(bx) is -(be^{4x}cos(bx))/(b^2+16)+(4e^{4x}sin(bx))/(b^2+16)+C

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integral of e^{4x}sin(bx)

\int e^{4 x} sin (b x) d x

- \frac{b e ^{4 x} cos ( b x )}{b ^{2} + 16} + \frac{4 e ^{4 x} sin ( b x )}{b ^{2} + 16} + C

Solution steps

\int e^{4 x} sin (b x) d x

Apply Integration By Parts

: - e^{4 x} \frac{1}{b} cos (b x) - \int - e^{4 x} \frac{4}{b} cos (b x) d x

= - e^{4 x} \frac{1}{b} cos (b x) - \int - e^{4 x} \frac{4}{b} cos (b x) d x

Take the constant out:

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

= - e^{4 x} \frac{1}{b} cos (b x) - (- \frac{4}{b} \cdot \int e^{4 x} cos (b x) d x)

Apply Integration By Parts

: e^{4 x} \frac{1}{b} sin (b x) - \int e^{4 x} \frac{4}{b} sin (b x) d x

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

Take the constant out:

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

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

Therefore

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

Isolate

\int e^{4 x} sin (b x) d x

= - \frac{b e ^{4 x} cos ( b x )}{b ^{2} + 16} + \frac{4 e ^{4 x} sin ( b x )}{b ^{2} + 16}

Add a constant to the solution

= - \frac{b e ^{4 x} cos ( b x )}{b ^{2} + 16} + \frac{4 e ^{4 x} sin ( b x )}{b ^{2} + 16} + C

f (x) = x + 9

\int 5 x^{\frac{3}{2}} d x

x \to \infty lim ((x) (e^{\frac{1}{x}}))

t \to 1 lim (t + 4)

x \to 5 lim (- (x - 3)^{2})

What is the integral of e^{4x}sin(bx) ?
The integral of e^{4x}sin(bx) is -(be^{4x}cos(bx))/(b^2+16)+(4e^{4x}sin(bx))/(b^2+16)+C