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# Evaluate the following integrals:

Question:

Evaluate the following integrals:

$\int e^{2 x} \sin (3 x+1) d x$

Solution:

Let $I=\int e^{2 x} \sin (3 x+1) d x$

Now Integrating by parts choosing $\sin (3 x+1)$ as first function and $e^{2 x}$ as second function we get,

$I=\sin (3 x+1) \int e^{2 x} d x-\int\left(\frac{d}{d x} \sin (3 x+1) \int e^{2 x} d x\right) d x$

$I=\frac{e^{2 x}}{2} \sin (3 x+1)-\int \frac{3 e^{2 x}}{2} \cos (3 x+1) d x$

Now again integrating by parts by taking $\cos (3 x+1)$ as first function and $\mathrm{e}^{2 x}$ as second function we get,

$I=\frac{e^{2 x}}{2} \sin (3 x+1)-\left[\cos (3 x+1) \int \frac{3 e^{2 x}}{2} d x-\int \frac{3}{2}\left(\frac{d}{d x} \cos (3 x+1) \int e^{2 x} d x\right) d x\right.$

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

$\int e^{2 x} \sin (3 x+1) d x=I$

Therefore,

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

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

$\frac{13 I}{4}=\frac{e^{2 x}}{2} \sin (3 x+1)-\frac{3}{4} e^{2 x} \cos (3 x+1)$

$I=\frac{e^{2 x}}{13}\{2 \sin (3 x+1)-3 \cos (3 x+1)\}+c$