Problems on Techniques of Integration

Use the Integration by Parts. Set

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

Then

$$\left\{\begin{array}{lll} du &=& -a \sin(ax) dx\\ v &=& \displaystyle \frac{1}{b} e^{bx}\;. \end{array}\right.$$

So

$$\begin{array}{lll} \displaystyle \int \cos(ax)e^{bx} dx &=&\d... ...os(ax)e^{bx} + \frac{a}{b}\int \sin(ax)e^{bx} dx \end{array}\;.$$

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

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

Then

$$\left\{\begin{array}{lll} du &=&a \cos(ax)dx\\ v &=&\displaystyle \frac{1}{b} e^{bx}\;. \end{array}\right.$$

So

$$\begin{array}{lll} \displaystyle \int \sin(ax)e^{bx} dx &=&\d... ... x)e^{b x} - \frac{a}{b}\int \cos(a x)e^{b x} dx \end{array}\;.$$

Hence

$$\int \cos(a x)e^{b x} dx = \frac{1}{b}\cos(a x)e^{b x} + \fra... ...in(a x)e^{b x} - \frac{a}{b}\int \cos(a x)e^{b x} dx\right] \;.$$

which implies

$$\left(1 + \frac{a^2 }{b^2 }\right) \int \cos(a x)e^{b x} dx = \frac{1}{b}\cos(a x)e^{b x} + \frac{a}{b^2 }\sin(ax)e^{b x}$$

or

$$\int \cos(ax)e^{bx} dx = \frac{b}{a^2 + b^2 } \cos(a x)e^{b x} + \frac{a}{a^2 + b^2 } \sin(a x)e^{b x} + C\;.$$

Detailed Answer.

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