Calculating integral?

Question

Hi,

I need to calculate the integral of:

-2sin(t) * e^(2cos(t)) + cos(t) * e^(sin(t)) dt.

Is it just: 2cos(t) * e^(2cos(t)) + sin(t) * e^(sin(t))?


Thanks!

Answer 1

No, it's not. If you differentiated your answer- remember that you have to use the product rule!- it's not the same as your original problem. You need to do a u substitution:
u=2cos(t)
du = -2sin(t) dt
that will give you the integral of e^u du, which is just e^u. Sub 2 cos(t) back in for u, and the answer is e^(2cos(t)), for the first part of the problem. Do the same procedure for the two terms after the plus sign, then add then together for your final answer.

Answer 2

Sorry.

Note, in the first term, that -2sin(t) is the derivative of the exponent of e^(2cos(t)).

So if you use the substitution u = 2cos(t), then du = -2sin(t)dt and you get integral of e^u du.

Similarly, in the second term, cos(t) is the derivative of the exponent of e^(sin(t))

So the answer is e^(2cos(t)) + e^(sin(t)) + C

Answer 3

∫ -2 sint * e^(2 cos(t) + cos(t)* e^(sin(t) dt

=>∫-2 sint * e^(2 cos(t) dt + ∫cos(t)* e^(sin(t) dt

first integral put 2 cost = x : - 2sint dt = dx

for second integral sint = y : cost dt = dy

=>∫e^x dx + ∫e^y dy

=> e^x + e^y + c

=> e^(2cos(t)) + e^(sin(t)) + c