(a) Compute the indefinite integral (find the antiderivatives) of Jear cos(4x) dx. (b) Write down the complete set of parts required to evaluate the above indefinite integral by integration by parts. (c) Write down the general formula or just the formula number of the Stewart's table of integration that corresponds to the above indefinite integral. (Note that the table of integration provided on huskyct follows the same nmbering of the Stewart's book.) (d) Apply the Fundamental Theorem of Calculus to evaluate eax cos(4x) dx.

Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
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Chapter2: Second-order Linear Odes
Section: Chapter Questions
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(a) Compute the indefinite integral (find the antiderivatives) of
| e3* cos(4x) dx.
(b) Write down the complete set of parts required to evaluate the above indefinite
integral by integration by parts.
(c) Write down the general formula or just the formula mumber of the Stewart's table
of integration that corresponds to the above indefinite integral. (Note that the table
of integration provided on huskyct follows the same mumbering of the Stewart's book.)
(d) Apply the Fundamental Theorem of Calculus to evaluate
e ax cos(4x) dx.
Transcribed Image Text:(a) Compute the indefinite integral (find the antiderivatives) of | e3* cos(4x) dx. (b) Write down the complete set of parts required to evaluate the above indefinite integral by integration by parts. (c) Write down the general formula or just the formula mumber of the Stewart's table of integration that corresponds to the above indefinite integral. (Note that the table of integration provided on huskyct follows the same mumbering of the Stewart's book.) (d) Apply the Fundamental Theorem of Calculus to evaluate e ax cos(4x) dx.
Expert Solution
Step 1

(a) We know that eaxcosbx+cdx=eaxa2+b2acosbx+c+bsinbx+c+C.

Now putting a=3, b=4 and c=0, we get anti-derivative or indefinite integral of the given integral such as

e3xcos4xdx=e3x253cos4x+4sin4x+C

 

(b) Let I=e3xcos4xdx, now by Integration by parts, we get 

I=e3xcos4xdx =e3xsin4x4-3e3xsin4x4dx=e3xsin4x4-34e3xsin4xdx=e3xsin4x4-34e3x-cos4x4-3e3x-cos4x4dx=e3xsin4x4+316e3xcos4x-916e3xcos4xdxI==e3xsin4x4+316e3xcos4x-916I1+916I=e3xsin4x4+316e3xcos4x2516I=e3x164sin(4x)+3cos4xI=e3x254sin(4x)+3cos4x

Hence, e3xcos4xdx=e3x254sin4x+3cos4x+C

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