What is the integral of e^{3z}cos(4z) ?

The integral of e^{3z}cos(4z) is (4e^{3z}sin(4z))/(25)+(3e^{3z}cos(4z))/(25)+C

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integral of e^{3z}cos(4z)

\int e^{3 z} cos (4 z) d z

\frac{4 e ^{3 z} sin ( 4 z )}{25} + \frac{3 e ^{3 z} cos ( 4 z )}{25} + C

Solution steps

\int e^{3 z} cos (4 z) d z

Apply Integration By Parts

: \frac{1}{4} e^{3 z} sin (4 z) - \int \frac{3}{4} e^{3 z} sin (4 z) d z

= \frac{1}{4} e^{3 z} sin (4 z) - \int \frac{3}{4} e^{3 z} sin (4 z) d z

Take the constant out:

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

= \frac{1}{4} e^{3 z} sin (4 z) - \frac{3}{4} \cdot \int e^{3 z} sin (4 z) d z

Apply Integration By Parts

: - \frac{1}{4} e^{3 z} cos (4 z) - \int - \frac{3}{4} e^{3 z} cos (4 z) d z

= \frac{1}{4} e^{3 z} sin (4 z) - \frac{3}{4} (- \frac{1}{4} e^{3 z} cos (4 z) - \int - \frac{3}{4} e^{3 z} cos (4 z) d z)

Take the constant out:

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

= \frac{1}{4} e^{3 z} sin (4 z) - \frac{3}{4} (- \frac{1}{4} e^{3 z} cos (4 z) - (- \frac{3}{4} \cdot \int e^{3 z} cos (4 z) d z))

Therefore

\int e^{3 z} cos (4 z) d z = \frac{1}{4} e^{3 z} sin (4 z) - \frac{3}{4} (- \frac{1}{4} e^{3 z} cos (4 z) - (- \frac{3}{4} \cdot \int e^{3 z} cos (4 z) d z))

Isolate

\int e^{3 z} cos (4 z) d z

= \frac{4 e ^{3 z} sin ( 4 z )}{25} + \frac{3 e ^{3 z} cos ( 4 z )}{25}

Add a constant to the solution

= \frac{4 e ^{3 z} sin ( 4 z )}{25} + \frac{3 e ^{3 z} cos ( 4 z )}{25} + C

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

\frac{d}{d x} (2 x^{3 π})

sin^{3} (2 t)

f (x) = - 3 x^{2} + 4 x - 1

x \to 2 lim (\frac{( x ^{2} - 5 x + 4 )}{x - 2})

What is the integral of e^{3z}cos(4z) ?
The integral of e^{3z}cos(4z) is (4e^{3z}sin(4z))/(25)+(3e^{3z}cos(4z))/(25)+C