In this question you will use integration by parts twice. Consider the integral

Question

I have asked this question before but maybe I did not provide all the questions so maybe there is some mistake.

In this question, you will use integration by parts twice. Consider the integral

I = ∫sin(4x)cosh(3x)dx

Think of this as s ∫uv′ dx Let u= cosh(3 x) .

(a) Enter v′

Answer: sin(4x)

(b) Enter u′

Answer: 3sinh(3x)

(c) Enter v (Note: you don’t need a constant of integration here.)

Answer: -1/4cos(4x)

Hence write the integral in the form I=uv−∫vu′dx:

(d) Enter uv

Answer: -1/4cosh(3x)*cos(4x)

(e) Enter vu′

Answer: -1/4cos(4x)*3sinh(3x)

Now apply integration by parts to this second integral, so consider the second integral as 

∫u2v′2dx. Note that any constants are included inside the integral. If u2= sinh(3 x) enter

(f) v′2

(g) u′2

Answer: 3cosh(3x)

(h) v2 (Note: you still don’t need a constant of integration here.)

(i) Hence write the integral in the form  I=uv−u2v2+∫v2u′2dx

I= () + ∫ () dx

(j) You should notice that the integral in (i) is a multiple of the original integral, I.

Hence rearrange the equation in (i) to determine the answer to the original integral, I.

(Remember you should include a constant of integration here, +c)

I=

I get (a),(b),(c),(d),(e),and (g) correct but I not sure how to do others

Answer 1

desmos.com/calculator/u4uvuzazgd

Second time for answering this question (first time used incorrect integrand).

Answer 2

This is really much easier to do using the D-I method

D I

+ cosh(3x) sin(4x)

- 3sinh(3x) -cos(4x)/4

+ 9cosh(3x) -sin(4x)/16

I = -cos(4x)cosh(3x)/4 + 3sin(4x)sinh(3x)/16 - 9sin(x)cos(3x)/16

= -cos(4x)cosh(3x)/4 + 3sin(4x)sinh(3x)/16 - 9I/16

Add -9I/16 to both sides

(1+9/16)I = -4cos(4x)cosh(3x)/16 + 3sin(4x)sinh(3x)/16

Solve for I

I = (-4cos(4x)cosh(3x) + 3sin(4x)sinh(3x))/25 + C

So, translating to uv-∫vdu method

u = cosh(3x)

u' = 3sinh(3x)

u'' = 9cosh(3x)

dv = sin(4x)

v = -cos(4x)/4

∫v = -sin(4x)/16

I = -cosh(4x)cos(3x)/4 +3sin(4x)sinh(3x)/16 -9I

...and then solve for I