Question

Given, I = e^(ax) cos(bx+c)

Integrate the following using:
i) Integration by parts
ii) Using differentiation
iii) Complex numbers
iv) Differential equation
v) Using Polar coordinates

Add new entry for each part.

Answer 1

Answer:

B is the correct answer ✔️

B) Br->Br> Br +

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Hope this helps u

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Answer 2

EXPLANATION.

(1) = Integration method.

∫eᵃˣ cos(bx + c).

As we know that,

in this type of question we can use integration by parts, we get.

If u and v are two function of x, then

∫(u v)dx = u(∫v dx) - ∫[du/dx . ∫v dx ]dx.

From the first letter of the word

I = Inverse trigonometric function.

L = Logarithmic function.

A = Algebraic function.

T = Trigonometric function.

E = Exponential function.

This is known as ILATE.

First arrange the function in the order according to letter of this word and then integration by parts.

Apply this rule in this question, we get.

⇒ cos(bx + c) = 1st function.

⇒ eᵃˣ = 2nd function.

$\sf \implies \displaystyle\int e^{ax} cos(bx + c).$

$\sf \implies I = cos(bx + c) \int e^{ax} - \displaystyle\int \bigg[\dfrac{d(cos bx + c)}{dx} . \int e^{ax} \bigg]dx$

$\sf \implies I = cos(bx + c) .\dfrac{e^{ax} }{a} \ - \displaystyle\int \bigg[-sin(bx + c)(b). \dfrac{e^{ax} }{a} \bigg]dx$

$\sf \implies I = cos(bx + c).\dfrac{e^{ax} }{a} \ + \dfrac{b}{a} \displaystyle\int \bigg[sin(bx + c) .e^{ax} \bigg]dx$

Again apply integration by parts, we get.

⇒ sin(bx + c) = 1st functions.

⇒ eᵃˣ = 2nd function.

$\sf \implies I = cos(bx + c).\dfrac{e^{ax} }{a} \ + \dfrac{b}{a} \bigg[sin(bx + c).\int e^{ax}\bigg] - \displaystyle\int \bigg[\dfrac{d(sin bx + c)}{dx} . \int e^{ax} \bigg]dx$

$\sf \implies I = cos(bx + c).\dfrac{e^{ax} }{a} \ + \dfrac{b}{a} \bigg[sin(bx + c).\dfrac{e^{ax} }{a} \bigg] - \displaystyle\int\bigg[cos(bx + c)(b) .\dfrac{e^{ax} }{a} \bigg]dx.$

$\sf \implies I = cos(bx + c).\dfrac{e^{ax} }{a} \ + \dfrac{b}{a^{2} }\bigg[sin(bx + c).e^{ax} \bigg] - \dfrac{b^{2} }{a^{2} } \displaystyle\int cos(bx + c) .e^{ax}$

$\sf \implies I = cos(bx + c).\dfrac{e^{ax} }{a} \ + \dfrac{b}{a^{2} }\bigg[sin(bx + c) .e^{ax} \bigg] - \dfrac{b^{2} }{a^{2} } (I)$

$\sf \implies I \ + \dfrac{b^{2}I }{a^{2} } = cos(bx + c).\dfrac{e^{ax} }{a} + \dfrac{b}{a^{2} } sin(bx + c).e^{ax} .$

$\sf \implies \dfrac{(a^{2} + b^{2})I }{a^{2} } = \dfrac{e^{ax} }{a} \bigg[cos(bx + c) + \dfrac{b sin(bx + c)}{a} \bigg].$

$\sf \implies \dfrac{(a^{2} + b^{2})I }{a^{2} } = \dfrac{e^{ax} }{a} \bigg[\dfrac{acos(bx + c) + bsin(bx + c)}{a} \bigg].$

$\sf \implies \dfrac{(a^{2} + b^{2})I }{a^{2} } = \dfrac{e^{ax} }{a^{2} } \bigg[acos(bx + c) + bsin(bx + c) \bigg].$

$\sf \implies (a^{2} + b^{2} )I = e^{ax} \bigg[acos(bx + c) + bsin(bx + c) \bigg].$

$\sf \implies I = \dfrac{e^{ax} }{a^{2} + b^{2} } \bigg[acos(bx + c) + bsin(bx + c) \bigg] + C.$

$\sf \implies \displaystyle\int e^{ax} cos(bx + c) = \dfrac{e^{ax} }{a^{2} + b^{2} } \bigg[acos(bx + c) + bsin(bx + c) \bigg] + C.$

(2) = Differentiation method.

⇒ y = eᵃˣ cos(bx + c).

As we know that,

⇒ d[f(x).g(x)]/dx = f(x).d[g(x)]/dx + g(x).d[f(x)]/dx.

Using this rule in equation, we get.

⇒ dy/dx = eᵃˣ . d[cos(bx + c)]/dx + cos(bx + c).d[eᵃˣ]/dx.

⇒ dy/dx = eᵃˣ. -sin(bx + c)(b) + cos(bx + c). eᵃˣ(a).

⇒ dy/dx = acos(bx + c)eᵃˣ - eᵃˣ.sin(bx + c)(b).

⇒ dy/dx = eᵃˣ[acos(bx + c) - bsin(bx + c)].