Find the integral of (e^cosx)sinx

1 answer

how about -e^cosx + c

(differentiate it to test my answer)

Similar Questions

I ran across this one the other day. See what you can make of it.Find the integral of e<sup><sup>cosx</sup></sup>sinx -

Top answer: How about you ask someone to explain it and not give you the answer Read more.
Simplify #3:[cosx-sin(90-x)sinx]/[cosx-cos(180-x)tanx] = [cosx-(sin90cosx-cos90sinx)sinx]/[cosx-(cos180cosx+sinx180sinx)tanx] =

Top answer: looks good to me Read more.
Use the symmetry of the graphs of the sine and cosine functions as an aid in evaluating each definite integral.(a) Integral of

Top answer: To use the symmetry of the graphs of the sine and cosine functions in evaluating definite integrals, Read more.
If y=3/(sinx+cosx) , find dy/dxA. 3sinx-3cosx B. 3/(sinx+cosx)^2 C. -3/(sinx+cosx)^2 D. 3(cosx-sinx)/(sinx+cosx)^2 E.

Top answer: ever heard of the chain rule? y = 3/u y' = -3/u^2 u' Now just plug in u = sinx+cosx Read more.
Prove that each of these equations is an identity.A) (1 + sinx + cos x)/(1 + sinx + cosx)=(1 + cosx)/sinx B) (1 + sinx +

Top answer: you must have a typo in A since the left side equals 1 for B multiply the left side by Read more.
Find the derivative of the function.y = integral from cosx to sinx (ln(8+3v)) dv lower limit = cosx upper limit = sinx y'(x) =

Top answer: Just use Lebniz's Rule: d/dx ∫[a(x),b(x)] f(t) dt = f(b(x))b' - f(a(x))a' d/dx ∫[cosx,sinx] Read more.
Evaluate the following integralintegral 1 = a and b = 4 of sinx dx/(1+cos)^2 u = cosx du = -sinx dx so from here I don't know if

Top answer: u = 1 + cos x du = -sin x dx so we have -du/u Read more.
tanx+secx=2cosx(sinx/cosx)+ (1/cosx)=2cosx (sinx+1)/cosx =2cosx multiplying both sides by cosx sinx + 1 =2cos^2x sinx+1 =

Top answer: The mistake occurred when you simplified the equation (sinx+1)/cosx = 2cosx to sinx + 1 = 2cos^2x. Read more.
I'm working this problem:∫ [1-tan^2 (x)] / [sec^2 (x)] dx ∫(1/secx)-[(sin^2x/cos^2x)/(1/cosx) ∫cosx-sinx(sinx/cosx)

Top answer: I think your 2nd line is bogus, and the brackets are unbalanced. (1-tan^2)/sec^2 = 1/sec^2 - Read more.
Trigonometric IdentitiesProve: (tanx + secx -1)/(tanx - secx + 1)= tanx + secx My work so far: (sinx/cosx + 1/cosx +

Top answer: To continue proving the trigonometric identity, we can simplify the expression obtained so far: Read more.