Question

Evaluate : ∫e2xsinxdx

Answer 1

Answer:

Hence,

$I=\int\limits {e^{2x}sinx} \, dx=\dfrac{2}{5}\sin x.e^{2x}-\dfrac{1}{5}\cos x.e^{2x}$

Step-by-step explanation:

We have to evaluate the integral:

$I=\int\limits {e^{2x}sinx} \, dx$

We will use the integration by parts we know that the trignometric function is considered as the first function and exponential function as the second to obtain,

$I=\sin x \dfrac{e^{2x}}{2}-\int {\cos x(\dfrac{e^{2x}}{2}}) \, dx \\\\I=\sin x \dfrac{e^{2x}}{2}-\dfrac{\cos x}{2}.(\dfrac{e^2x}{2})+\int {\dfrac{1}{2}\times (-\sin x)(\dfrac{e^2x}{2})} \, dx \\\\\\I=\sin x \dfrac{e^{2x}}{2}-\dfrac{\cos x}{4}e^{2x}-\dfrac{1}{4}\int {\sin x .e^{2x}} \, dx \\\\I=\sin x \dfrac{e^{2x}}{2}-\dfrac{\cos x}{4}e^{2x}-\dfrac{I}{4}\\\\I+\dfrac{I}{4}=\dfrac{1}{2}\sin x.e^{2x}-\dfrac{1}{4}\cos x\e^{2x}\\\\\dfrac{5}{4}I=\dfrac{1}{2}\sin x.e^{2x}-\dfrac{1}{4}\cos x.e^{2x}$

$I=\dfrac{2}{5}\sin x.e^{2x}-\dfrac{1}{5}\cos x.e^{2x}$

Hence,

$I=\int\limits {e^{2x}sinx} \, dx=\dfrac{2}{5}\sin x.e^{2x}-\dfrac{1}{5}\cos x.e^{2x}$