Question

Integrate the following problem:
$\int {e^{-x} \cdot cos(2x)} \, dx$​

Answer 1

$\huge{\underline{\underline{\mathcal\color{blue}{\bigstar {Answer}}}}}$

$\red{\displaystyle\int e^{-x}\cos(2x)\, dx=\dfrac{2\sin(2x)-\cos(2x)}{5e^x}+C}$

Step-by-step explanation:

$\rightsquigarrow$ We would like to integrate the following integral:

$\bold{\displaystyle \int e^{-x}\cdot \cos(2x)\, dx}$

$\rightsquigarrow$ Since this is a product of two functions, we can consider using Integration by Parts given by:

$\displaystyle \int u\, dv =uv-\int v\, du$

$\rightsquigarrow$ So, let’s choose our u and dv. We can choose u base on the following guidelines: LIATE; or, logarithmic, inverse trig., algebraic, trigonometric, and exponential.

$\rightsquigarrow$ Since trigonometric comes before exponential, we will let:

$u=\cos(2x)\text{ and } dv=e^{-x}\, dx$

$\rightsquigarrow$ By finding the differential of the left and integrating the right, we acquire:

$du=-2\sin(2x)\text{ and } v=-e^{-x}$

$\rightsquigarrow$ So, our integral becomes:

$\displaystyle \int e^{-x}\cdot \cos(2x)\, dx=(\cos(2x))(-e^{-x})-\int (-e^{-x})(-2\sin(2x))\, dx$

$\rightsquigarrow$ Simplify:

$\displaystyle \int e^{-x}\cdot \cos(2x)\, dx=-e^{-x}\cos(2x)-2\int e^{-x}\sin(2x)\, dx$

$\rightsquigarrow$ Since we ended up with another integral of a product of two functions, we can apply integration by parts again. Using the above guidelines, we get that:

$u=\sin(2x)\text{ and } dv=e^{-x}\, dx$

$\rightsquigarrow$ By finding the differential of the left and integrating the right, we acquire:

$du=2\cos(2x)\, dx\text{ and } v=-e^{-x}$

$\rightsquigarrow$ This yields:

$\displaystyle \int e^{-x}\cdot \cos(2x)\, dx=-e^{-x}\cos(2x)-2\Big[(\sin(2x))(-e^{-x})-\int (-e^{-x})(2\sin(2x))\, dx\Big]$

$\rightsquigarrow$ Simplify:

$\displaystyle \int e^{-x}\cdot \cos(2x)\, dx=-e^{-x}\cos(2x)-2\Big[-e^{-x}\sin(2x)+2\int e^{-x}\cos(2x)\, dx\Big]$

$\rightsquigarrow$ We can distribute:

$\displaystyle \int e^{-x}\cdot \cos(2x)\, dx=-e^{-x}\cos(2x)+2e^{-x}\sin(2x)-4\int e^{-x}\cos(2x)\, dx$

$\rightsquigarrow$ The integral on the right is the same as our original integral. So, we can isolate it:

$\displaystyle \Big(\int e^{-x}\cos(2x)\, dx\Big)+4\Big(\int e^{-x}\cos(2x)\, dx)\Big)=-e^{-x}\cos(2x)+2e^{-x}\sin(2x)$

$\rightsquigarrow$ Combine like integrals:

$\displaystyle 5 \int e^{-x}\cos(2x)\, dx=-e^{-x}\cos(2x)+2e^{-x}\sin(2x)$

$\rightsquigarrow$ We can factor out an e⁻ˣ from the right:

$\displaystyle 5\int e^{-x}\cos(2x)\, dx=e^{-x}\Big(-\cos(2x)+2\sin(2x)\Big)$

$\rightsquigarrow$ Dividing both sides by 5 yields:

$\displaystyle \int e^{-x}\cos(2x)\, dx=\dfrac{e^{-x}}{5}\Big(-\cos(2x)+2\sin(2x)\Big)$

$\hookrightarrow$ Rewrite. We of course also need the constant of integration. Therefore, our final answer is:

$\displaystyle\int e^{-x}\cos(2x)\, dx=\dfrac{2\sin(2x)-\cos(2x)}{5e^x}+C$

Answer 2

EXPLANATION.

∫e⁻ˣ.cos(2x)dx.

In this question, we can use ILATE Formula.

$\sf \implies \int (uv)dx = u \int vdx - \int \Bigg( (\dfrac{du}{dx}) \int vdx \Bigg)dx.$

$\sf \implies \int (u.v)dx = u. (\int v) - \int \Bigg( (u'). \int v \Bigg) dx.$

ILATE =.

I = Inverse Trigonometric Function.

L = Logarithmic Function.

A = Algebraic Function.

T = Trigonometric Function.

E = Exponential Function.

Always move from left to right, just like.

I ⇒ L ⇒ A ⇒ T ⇒ E.

⇒ ∫e⁻ˣ.cos(2x)dx.

As we know that,

e⁻ˣ = Exponential function.

Cos(2x) = Trigonometric Function.

Therefore,

e⁻ˣ is considered as (2)nd Function.

Cos(2x) is considered as (1)st Function.

$\sf \implies -Cos(2x).e^{-x} - \int (2sin(2x) x. e^{-x})dx.$

Again, we can apply ILATE in (sin(x).e⁻ˣ we get,

$\sf \implies -cos(2x).ex^{-x} + \int(2sin(2x).e^{-x} )dx.$

$\sf \implies \int-e^{-x}Cos(2x) + 2\int(e ^{-x}Sin(2x)dx.$

$\sf \implies I = \int-e^{-x}Cos(2x) + 2\int(e ^{-x}Sin(2x)dx.$

Let I₂ = ∫e⁻ˣsin2xdx.

I₂ = Sin(2x)e⁻ˣ - ∫ 2sin2x.e⁻ˣdx.

I = Sin(2x)e⁻ˣ - 2I.

I = e⁻ˣcos(2x) - 2e⁻ˣsin2x - 4I.

5I = e⁻ˣcos2x - 2e⁻ˣsin2x + c.

I = e⁻ˣ[ cos2x - 2sin2x]/5 + c.