This is really Hard Whats is the Integral to e^x cos(cx) dx where c is a constant?

Question

Ive done it but I keep getting 0=0 or the integral of (e^x cos(cx) )= e^x cos(cx) and that would imply c = 0

Answer 1

∫e^x cos(cx) dx



integrate by parts twice



u = cos(cx) ===> du = - c sin(cx) dx

dv = e^x dx ===> v = e^x



∫e^x cos(cx) dx = e^x cos(cx) + ∫c e^x sin(cx) dx



again integrate by parts



u = c sin(cx) ===> du = c^2 cos(cx) dx

dv = e^x dx ===> v = e^x



∫e^x cos(cx) dx = e^x cos(cx) + [c e^x sin(cx) - ∫c^2 e^x cos (cx) dx ]



∫e^x cos(cx) dx = e^x cos(cx) + c e^x sin(cx) - ∫c^2 e^x cos (cx) dx



(1 + c^2) ∫e^x cos(cx) dx = e^x cos(cx) + c e^x sin(cx)



∫e^x cos(cx) dx = e^x [ cos(cx) + c sin(cx) ] / (1 + c^2 )

Answer 2

You're probably mostly on the right track; however, if you choose u = e^x for the first IBP, you will need to pick it for the second round in order for the IBPs to work in your favor. It's similar if you picked the trigonometric function as u. You MUST be consistent with your choice!



For the first round of IBP, let:



u = e^x ==> du = e^x dx

dv = cos(cx) dx ==> v = (1/c)sin(cx).



Then:



∫ e^x*cos(cx) dx

=> uv - ∫ v du

= (1/c)e^x*sin(cx) - ∫ (1/c)e^x*sin(cx) dx

= (1/c)e^x*sin(cx) - 1/c ∫ e^x*sin(cx) dx.



Now, we let:



u = e^x ==> du = e^x dx

dv = sin(cx) dx ==> v = (-1/c)cos(cx)

(Notice how I picked e^x as u and the trig function as dv?)



By a second round of IBP:



∫ e^x*cos(cx) dx

= (1/c)e^x*sin(cx) - 1/c ∫ e^x*sin(cx) dx

=> (1/c)e^x*sin(cx) - (1/c)(uv - ∫ v du)

= (1/c)e^x*sin(cx) - (1/c)[(-1/c)e^x*cos(cx) - ∫ (-1/c)e^x*cos(cx) dx]

= (1/c)e^x*sin(cx) - [(-1/c^2)e^x*cos(cx) + 1/c^2 ∫ e^x*cos(cx) dx]

= (1/c)e^x*sin(cx) + (1/c^2)e^x*cos(cx) - 1/c^2 ∫ e^x*cos(cx) dx.



With I = ∫ e^x*cos(cx) dx, we have:



I = (1/c)e^x*sin(cx) + (1/c^2)e^x*cos(cx) - (1/c^2)I

==> [(c^2 + 1)/c^2]I = (1/c)e^x*sin(cx) + (1/c^2)e^x*cos(cx)

==> I = c^2/(c^2 + 1) * [(1/c)e^x*sin(cx) + (1/c^2)e^x*cos(cx)] + C

==> ∫ e^x*cos(cx) dx = e^x * [c*sin(cx) + cos(cx)]/(c^2 + 1) + C.



I hope this helps!

Answer 3

(e^(x)cos(cx))=e^(x)cos(cx)



Differentiate both sides of the equation ((e^(x)cos(cx))=e^(x)cos(cx)).

(d)/(dc) ((e^(x)cos(cx)))=(d)/(dc) (e^(x)cos(cx))



Find the derivative of the expression.

(d)/(dc) (e^(x)cos(cx))



Find the derivative of the expression.

(d)/(dc) e^(x)cos(cx)



The chain rule states that the derivative of a composite function (f o g)' is equal to (f' o g)*g'. To find the derivative of e^(x)cos(cx), find the derivatives of each portion of the function and use the chain rule formula.

(d)/(du) e^(x)cos(u)*(d)/(dc) cx



The derivative of e^(x)cos(u) is -(e^(x)sin(u)).

(d)/(du) e^(x)cos(u)=-(e^(x)sin(u))



Multiply -1 by the e^(x)sin(u) inside the parentheses.

(d)/(du) e^(x)cos(u)=-e^(x)sin(u)



Use the product rule to find the derivative of cx. The product rule states that (fg)'=f'g+fg'.

[(d)/(dc) c](x)+(c)[(d)/(dc) x]



The derivative of c is (dx)/(dc) c=1.

(dx)/(dc) c=1



Substitute the derivative back into the product rule formula.

(d)/(dc) cx=(1)(x)+(c)[(d)/(dc) x]



The derivative of x is (dx)/(dc) x=x'.

(dx)/(dc) x=x'



Substitute the derivative back into the product rule formula.

(d)/(dc) cx=(1)(x)+(c)(x')



Simplify the derivative.

(d)/(dc) cx=x+x'c



Replace the variable u with cx in the expression.

(d)/(du) e^(x)cos(u)=-e^(x)sin((cx))



Replace the variable u with cx in the expression.

x+x'c



Form the derivative by substituting the values for each portion into the chain rule formula.

=-e^(x)sin((cx))*(x+x'c)



Remove the parentheses around the expression -e^(x)xsin((cx))-x'ce^(x)sin((cx)).

(d)/(dc) e^(x)cos(cx)=-e^(x)xsin((cx))-x'ce^(x)sin((cx))



Find the derivative of the expression.

(d)/(dc) e^(x)cos(cx)



The chain rule states that the derivative of a composite function (f o g)' is equal to (f' o g)*g'. To find the derivative of e^(x)cos(cx), find the derivatives of each portion of the function and use the chain rule formula.

(d)/(du) e^(x)cos(u)*(d)/(dc) cx



The derivative of e^(x)cos(u) is -(e^(x)sin(u)).

(d)/(du) e^(x)cos(u)=-(e^(x)sin(u))



Multiply -1 by the e^(x)sin(u) inside the parentheses.

(d)/(du) e^(x)cos(u)=-e^(x)sin(u)



Use the product rule to find the derivative of cx. The product rule states that (fg)'=f'g+fg'.

[(d)/(dc) c](x)+(c)[(d)/(dc) x]



The derivative of c is (dx)/(dc) c=1.

(dx)/(dc) c=1



Substitute the derivative back into the product rule formula.

(d)/(dc) cx=(1)(x)+(c)[(d)/(dc) x]



The derivative of x is (dx)/(dc) x=x'.

(dx)/(dc) x=x'



Substitute the derivative back into the product rule formula.

(d)/(dc) cx=(1)(x)+(c)(x')



Simplify the derivative.

(d)/(dc) cx=x+x'c



Replace the variable u with cx in the expression.

(d)/(du) e^(x)cos(u)=-e^(x)sin((cx))



Replace the variable u with cx in the expression.

x+x'c



Form the derivative by substituting the values for each portion into the chain rule formula.

=-e^(x)sin((cx))*(x+x'c)



Remove the parentheses around the expression -e^(x)xsin((cx))-x'ce^(x)sin((cx)).

(d)/(dc) e^(x)cos(cx)=-e^(x)xsin((cx))-x'ce^(x)sin((cx))



Form the equation involving x'

-e^(x)xsin((cx))-x'ce^(x)sin((cx))=-e^(x)xsin((cx))-x'ce^(x)sin((cx))



Remove the parentheses around the expression cx.

-e^(x)xsin((cx))-x'ce^(x)sin((cx))=-e^(x)x(sin(cx))-x'ce^(x)sin((cx))



Remove the extra parentheses around the factors.

-e^(x)xsin((cx))-x'ce^(x)sin((cx))=-e^(x)xsin(cx)-x'ce^(x)sin((cx))



Remove the parentheses around the expression cx.

-e^(x)xsin((cx))-x'ce^(x)sin((cx))=-e^(x)xsin(cx)-x'ce^(x)(sin(cx))



Remove the extra parentheses around the factors.

-e^(x)xsin((cx))-x'ce^(x)sin((cx))=-e^(x)xsin(cx)-x'ce^(x)sin(cx)



Remove the parentheses around the expression cx.

-e^(x)x(sin(cx))-x'ce^(x)sin((cx))=-e^(x)xsin(cx)-x'ce^(x)sin(cx)



Remove the extra parentheses around the factors.

-e^(x)xsin(cx)-x'ce^(x)sin((cx))=-e^(x)xsin(cx)-x'ce^(x)sin(cx)



Remove the parentheses around the expression cx.

-e^(x)xsin(cx)-x'ce^(x)(sin(cx))=-e^(x)xsin(cx)-x'ce^(x)sin(cx)



Remove the extra parentheses around the factors.

-e^(x)xsin(cx)-x'ce^(x)sin(cx)=-e^(x)xsin(cx)-x'ce^(x)sin(cx)



Since -x'ce^(x)sin(cx) contains the variable to solve for, move it to the left-hand side of the equation by adding x'ce^(x)sin(cx) to both sides.

-e^(x)xsin(cx)-x'ce^(x)sin(cx)+x'ce^(x)sin(cx)=-e^(x)xsin(cx)



Since -x'ce^(x)sin(cx) and x'ce^(x)sin(cx) are like terms, subtract x'ce^(x)sin(cx) from -x'ce^(x)sin(cx) to get 0.

-e^(x)xsin(cx)=-e^(x)xsin(cx)



Since the bases are the same, then two expressions are only equal if the exponents are also equal.

x=x



Since x contains the variable to solve for, move it to the left-hand side of the equation by subtracting x from both sides.

x-x=0



Since x and -x are like terms, add -x to x to get 0.

0=0



Since 0=0, the equation will always be true.

(dx)/(dc)=Alwx'ys True