What is the integral of 3e^{5x-2}cos(7-4x) ?

The integral of 3e^{5x-2}cos(7-4x) is 3(-(4e^{5x-2}sin(-4x+7))/(41)+(5e^{5x-2}cos(-4x+7))/(41))+C

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integral of 3e^{5x-2}cos(7-4x)

\int 3 e^{5 x - 2} cos (7 - 4 x) d x

3 (- \frac{4 e ^{5 x - 2} sin ( - 4 x + 7 )}{41} + \frac{5 e ^{5 x - 2} cos ( - 4 x + 7 )}{41}) + C

Solution steps

\int 3 e^{5 x - 2} cos (7 - 4 x) d x

Take the constant out:

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

= 3 \cdot \int e^{5 x - 2} cos (7 - 4 x) d x

Apply Integration By Parts

: - \frac{1}{4} e^{5 x - 2} sin (7 - 4 x) - \int - \frac{5}{4} e^{5 x - 2} sin (7 - 4 x) d x

= 3 (- \frac{1}{4} e^{5 x - 2} sin (7 - 4 x) - \int - \frac{5}{4} e^{5 x - 2} sin (7 - 4 x) d x)

Take the constant out:

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

= 3 (- \frac{1}{4} e^{5 x - 2} sin (7 - 4 x) - (- \frac{5}{4} \cdot \int e^{5 x - 2} sin (7 - 4 x) d x))

Apply Integration By Parts

: \frac{1}{4} e^{5 x - 2} cos (7 - 4 x) - \int \frac{5}{4} e^{5 x - 2} cos (7 - 4 x) d x

= 3 (- \frac{1}{4} e^{5 x - 2} sin (7 - 4 x) - (- \frac{5}{4} (\frac{1}{4} e^{5 x - 2} cos (7 - 4 x) - \int \frac{5}{4} e^{5 x - 2} cos (7 - 4 x) d x)))

Take the constant out:

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

= 3 (- \frac{1}{4} e^{5 x - 2} sin (7 - 4 x) - (- \frac{5}{4} (\frac{1}{4} e^{5 x - 2} cos (7 - 4 x) - \frac{5}{4} \cdot \int e^{5 x - 2} cos (7 - 4 x) d x)))

Therefore

3 \cdot \int e^{5 x - 2} cos (7 - 4 x) d x = 3 (- \frac{1}{4} e^{5 x - 2} sin (7 - 4 x) - (- \frac{5}{4} (\frac{1}{4} e^{5 x - 2} cos (7 - 4 x) - \frac{5}{4} \cdot \int e^{5 x - 2} cos (7 - 4 x) d x)))

Isolate

\int e^{5 x - 2} cos (7 - 4 x) d x

= 3 (- \frac{4 e ^{5 x - 2} sin ( - 4 x + 7 )}{41} + \frac{5 e ^{5 x - 2} cos ( - 4 x + 7 )}{41})

Add a constant to the solution

= 3 (- \frac{4 e ^{5 x - 2} sin ( - 4 x + 7 )}{41} + \frac{5 e ^{5 x - 2} cos ( - 4 x + 7 )}{41}) + C

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What is the integral of 3e^{5x-2}cos(7-4x) ?
The integral of 3e^{5x-2}cos(7-4x) is 3(-(4e^{5x-2}sin(-4x+7))/(41)+(5e^{5x-2}cos(-4x+7))/(41))+C