Problems on Techniques of Integration

Use the Integration by Parts. Set

$$\left\{\begin{array}{lll} u &=& \sin(2x)\\ dv &=& e^{3x} dx\;. \end{array}\right.$$

Then

$$\left\{\begin{array}{lll} du &=& 2 \cos(2x) dx\\ v &=& \displaystyle \frac{1}{3} e^{3x}\;. \end{array}\right.$$

So

$$\begin{array}{lll} \displaystyle \int \sin(2x)e^{3x} dx &=&\d... ...in(2x)e^{3x} - \frac{2}{3}\int \cos(2x)e^{3x} dx \end{array}\;.$$

In order to integrate the function $\cos(2x)e^{3x}$, we will need to do another integration by parts. Set

$$\left\{\begin{array}{lll} u &=&\cos(2x)\\ dv &=& e^{3x}dx\;. \end{array}\right.$$

Then

$$\left\{\begin{array}{lll} du &=&-2 \sin(2x)dx\\ v &=&\displaystyle \frac{1}{3} e^{3x}\;. \end{array}\right.$$

So

$$\begin{array}{lll} \displaystyle \int \cos(2x)e^{3x} dx &=&\d... ...os(2x)e^{3x} + \frac{2}{3}\int \sin(2x)e^{3x} dx \end{array}\;.$$

Hence

$$\int \sin(2x)e^{3x} dx = \frac{1}{3}\sin(2x)e^{3x} - \frac{2}... ...3}\cos(2x)e^{3x} + \frac{2}{3}\int \sin(2x)e^{3x} dx\right] \;.$$

which implies

$$\left(1 + \frac{4}{9}\right) \int \sin(2 x)e^{3 x} dx = \frac{1}{3}\sin(2x)e^{3x} - \frac{2}{9}\cos(2x)e^{3x}$$

or

$$\int \sin(2x)e^{3x} dx = \frac{3}{13} \sin(2x)e^{3x} - \frac{2}{13} \cos(2x)e^{3x} + C\;.$$

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