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Question: How do you integrate \({{e}^{3x}}\sin (2x)dx\) ?...

How do you integrate e3xsin⁡(2x)dx{{e}^{3x}}\sin (2x)dx ?

Explanation

Solution

To solve these types of questions use the concept and formula of Integration by Parts. Take the two functions separately and consider them as function 1 and function 2. Then apply the formula of Integration by Parts and simplify the expressions to get the final answer.

Complete step-by-step solution:
Given expression to integrate is:
e3xsin⁡(2x)dx{{e}^{3x}}\sin \left( 2x \right)dx
Let I=∫e3xsin⁡(2x)dxI=\int{{{e}^{3x}}\sin (2x)dx} …(i)(i)
We know that Integration by Parts can be done in the following way:
∫uvdx=u∫vdx−∫(dudx∫vdx)dx\int{uvdx=u\int{vdx-\int{\left( \dfrac{du}{dx}\int{vdx} \right)}}}dx , where uu and vv are the two functions. Applying the above formula to the equation (i)(i) , we get:
⇒I=e3x(−cos⁡2x2)−∫3e3x(−cos⁡2x2)dx\Rightarrow I={{e}^{3x}}\left( \dfrac{-\cos 2x}{2} \right)-\int{3{{e}^{3x}}}\left( \dfrac{-\cos 2x}{2} \right)dx
Taking the constants out of the integration side,
⇒I=−e3x(cos⁡2x2)+32∫e3xcos⁡2xdx\Rightarrow I=-{{e}^{3x}}\left( \dfrac{\cos 2x}{2} \right)+\dfrac{3}{2}\int{{{e}^{3x}}}\cos 2xdx
Applying Integration by Parts again in the second term of the above expression, we get:
⇒I=−e3x(cos⁡2x2)+32[e3x(sin⁡2x2)−∫3e3x(sin⁡2x2)dx]\Rightarrow I=-{{e}^{3x}}\left( \dfrac{\cos 2x}{2} \right)+\dfrac{3}{2}\left[ {{e}^{3x}}\left( \dfrac{\sin 2x}{2} \right)-\int{3{{e}^{3x}}\left( \dfrac{\sin 2x}{2} \right)}dx \right]
Taking the constants out of the integration side,
⇒I=−e3x(cos⁡2x2)+3e3xsin⁡2x4−94∫e3xsin⁡2xdx\Rightarrow I=-{{e}^{3x}}\left( \dfrac{\cos 2x}{2} \right)+\dfrac{3{{e}^{3x}}\sin 2x}{4}-\dfrac{9}{4}\int{{{e}^{3x}}\sin 2xdx}
In the above expression, we can see clearly that the last term is the same as the expression given to us in the question statement, so we will replace it by IIand then further solve the expression.
⇒I=−e3x(cos⁡2x2)+3e3xsin⁡2x4−94I\Rightarrow I=-{{e}^{3x}}\left( \dfrac{\cos 2x}{2} \right)+\dfrac{3{{e}^{3x}}\sin 2x}{4}-\dfrac{9}{4}I
Transposing the like terms on the left-hand side of the equation,
⇒I+94I=−e3xcos⁡2x2+3e3xsin⁡2x4\Rightarrow I+\dfrac{9}{4}I=\dfrac{-{{e}^{3x}}\cos 2x}{2}+\dfrac{3{{e}^{3x}}\sin 2x}{4}
Simplifying the above expression by taking L.C.M and common terms, we get,
⇒I(9+44)=e3x4(3sin⁡2x−2cos⁡2x)\Rightarrow I\left( \dfrac{9+4}{4} \right)=\dfrac{{{e}^{3x}}}{4}\left( 3\sin 2x-2\cos 2x \right)
On further simplification, we can see that, we get the answer as:
⇒I=e3x13(3sin⁡2x−2cos⁡2x)+C\Rightarrow I=\dfrac{{{e}^{3x}}}{13}\left( 3\sin 2x-2\cos 2x \right)+C
Hence, on integrating e3xsin⁡(2x)dx{{e}^{3x}}\sin (2x)dx by using integration by parts method we get the answer as:
I=e3x13(3sin⁡2x−2cos⁡2x)+CI=\dfrac{{{e}^{3x}}}{13}\left( 3\sin 2x-2\cos 2x \right)+C

Note: While solving these types of questions that include the integration by parts method, it is extremely important to choose the first and second functions carefully so that the solution does not become too lengthy and can be solved easily. If the functions are taken in the wrong order, then the solution will keep going with no definite answer.
Basic integration rules play a very important role while solving these questions.