Can you help me provethis integral? thanks!?

Question

integral e^(au) cos nu du = (e^au(acos nu + n sin nu))/(a^2+n^2) + C

Answer 1

its 5 to the 2ed power

Answer 2

Ok the way of doing this problem is by using the integration by parts method. This method is useful when you have different classes of functions in multiplication with each other, say in your case, the LHS has an exponential and a trignometric function in multiplication. You only get 5 classes of functions in general high school or even 1st year university maths. The method works by remembering the simple formula called : ILATE..which stands for "inverse trignometric", "logarithmic" , "algebraic", "trignometric", "exponential". You can expect to almost always get nothing more than 2 classes of functions at the one time. How ever you may get a question like integrate : xcos(x)e^2x; which has algebraic trignometric and exponential functions. With integration by parts you want to look at the question and see which function comes first comes first in ILATE. In your question you have cos(nu) as the first function and e^(au) as the second one. so this is how you solve it after numbering ur functions as 1 and 2.

[(first function as it is) x (integration of second function)] - integrationof[(differentiation of first function) x (integration of second function)]

so here we go since you have the concept clear.

i'll start from the 2nd step, the first step is just rewriting the question again...

1/a(cosnu x e^(au)) + n/a[integrationof(sinnu x e^(au))]

now the second term of the above equation also has the same ILATE situation redeveloping again so we apply Ilate once more. The minus sign got converted to plus because diff of cos is -sin.

1/a(cosnu x e^(au)) + n/a[[(sinnu x e^(au))/a] - n/a[integrationof(cosnu x e^(au)]

now lets call the LHS of your question with the name "I". If you carefully notice, the second subterm of the second major term in the above equation looks like the LHS. So the above equation now becomes.

1/a(cosnu x e^(au)) + n/(a^2)[sinnu x e^(au)] - (n/a x I)

now you can take the 3rd term to the LHS get

I[1 + n/a] = 1/a(cosnu x e^(au)) + n/(a^2)[sinnu x e^(au)]

rearranging the terms on both sides should give you your answer. You might want to write everything down on paper before trying to understand it because I couldn't put the proper integral signs on here.

hope you got it, good luck.