Question

Find the following integral :-
$\displaystyle \sf \int e^{2x}sin3x \ dx$

Answer 1

Topic :-

Indefinite Integration

To Solve :-

$\sf{I=\displaystyle \int e^{2x}\sin 3x\:dx}$

Concept Used :-

Integration by Parts

$\sf{\displaystyle \int uv \:dx = u\int v\:dx - \int \left(\dfrac{du}{dx}\int v\:dx \right)dx}$

where u and v are differentiable functions.

u and v are generally chosen as per the order of the letters in ILATE, where

I represents Inverse Trigonometric Function

L represents Logarithmic Function

A represents Algebraic Function

T represents Trigonometric Function and

E represents Exponential Function

Solution :-

$\sf{I=\displaystyle \int e^{2x}\sin 3x\:dx}$

$\sf{Taking\:u=\sin3x \:and\:v=e^{2x}\:as\:per\:ILATE.}$

Applying Integration by parts,

$\sf{I=\displaystyle\sin3x\int e^{2x}\:dx - \int \left(\dfrac{d(\sin 3x)}{dx}\int e^{2x}\:dx \right)dx}$

$\sf{I=\displaystyle\sin3x\cdot\dfrac{e^{2x}}{2} - \int \left(\dfrac{d(\sin 3x)}{dx}\cdot\dfrac{e^{2x}}{2}\right)dx}$

$\sf{\left(\because\displaystyle \int e^{ax}\:dx=\dfrac{e^{ax}}{a}+C \right)}$

$\sf{\displaystyle I=\sin3x\cdot\dfrac{e^{2x}}{2} - \int \left(\cos3x\cdot\dfrac{d(3x)}{dx}\cdot\dfrac{e^{2x}}{2}\right)dx}$

$\sf{\left(\because \dfrac{d(\sin t)}{dx}=\cos t\cdot\dfrac{dt}{dx} \right)}$

$\sf{I = \displaystyle\sin3x\cdot\dfrac{e^{2x}}{2} - \int \left(3\cos3x\cdot\dfrac{e^{2x}}{2}\right)dx}$

$\sf{\left(\because \dfrac{d(kx)}{dx}=k,when\:k\:is\:a\:constant. \right)}$

$\sf{I=\displaystyle\sin3x\cdot\dfrac{e^{2x}}{2} - \dfrac{3}{2}\int \cos3x\cdot e^{2x}\:dx}$

$\sf{\left(\because \displaystyle \int k\cdot f(x)\:dx=k\int f(x)\:dx,when\:k\:is\:a\:constant. \right)}$

$\sf{Take\:\displaystyle\int \cos3x\cdot e^{2x}\:dx=J}$

$\sf{I=\displaystyle \sin3x\cdot\dfrac{e^{2x}}{2} - \dfrac{3}{2}J}$

Solving J,

$\sf{J=\displaystyle \int \cos3x\cdot e^{2x}\:dx}$

$\sf{Taking\:u=\cos3x \:and\:v=e^{2x}\:as\:per\:ILATE.}$

Applying Integration by parts,

$\sf {J=\displaystyle\cos3x\int e^{2x}\:dx - \int \left(\dfrac{d(\cos 3x)}{dx}\int e^{2x}\:dx \right)dx}$

$\sf{J=\displaystyle \cos3x\cdot\dfrac{e^{2x}}{2} - \int \left(\dfrac{d(\cos 3x)}{dx}\cdot\dfrac{e^{2x}}{2}\right)dx}$

$\sf{J=\displaystyle \cos3x\cdot\dfrac{e^{2x}}{2} - \int \left(-\sin3x\cdot\dfrac{d(3x)}{dx}\cdot\dfrac{e^{2x}}{2}\right)dx}$

$\sf{\left(\because \dfrac{d(\cos t)}{dx}=-\sin t\cdot\dfrac{dt}{dx} \right)}$

$\sf{J=\displaystyle \cos3x\cdot\dfrac{e^{2x}}{2} +\int \left(3\sin3x\cdot\dfrac{e^{2x}}{2}\right)dx}$

$\sf{\left(\because \dfrac{d(kx)}{dx}=k,when\:k\:is\:a\:constant. \right)}$

$\sf{J=\displaystyle \cos3x\cdot\dfrac{e^{2x}}{2} + \dfrac{3}{2}\int \sin3x\cdot e^{2x}\:dx}$

$\sf{J= \cos3x\cdot\dfrac{e^{2x}}{2} + \dfrac{3}{2}I}$

Substitute value of 'J' in 'I',

$\sf{I=\displaystyle \sin3x\cdot\dfrac{e^{2x}}{2} - \dfrac{3}{2}J}$

$\sf{I=\displaystyle \sin3x\cdot\dfrac{e^{2x}}{2} - \dfrac{3}{2}\left( \cos3x\cdot\dfrac{e^{2x}}{2} + \dfrac{3}{2}I \right)}$

$\sf{I=\displaystyle \dfrac{\sin3x\cdot e^{2x}}{2} - \dfrac{3\cos3x\cdot e^{2x}}{4} - \dfrac{9}{4}I}$

$\sf{I+\dfrac{9}{4}I=\displaystyle\dfrac{2\sin3x\cdot e^{2x}}{4} - \dfrac{3\cos3x\cdot e^{2x}}{4}}$

$\sf{ \dfrac{13}{4}I= \dfrac{2\sin3x\cdot e^{2x}-3\cos3x\cdot e^{2x}}{4}}$

$\sf{ 13I= 2\sin3x\cdot e^{2x}-3\cos3x\cdot e^{2x}}$

$\sf{ I= \dfrac{2\sin3x\cdot e^{2x}-3\cos3x\cdot e^{2x}}{13}+C}$

Answer :-

$\underline{\boxed{\sf{\displaystyle \int e^{2x}\sin 3x\:dx= \dfrac{2\sin3x\cdot e^{2x}-3\cos3x\cdot e^{2x}}{13}+C}}}$

Answer 2

Answer:

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