Ok so I just had a calculus final and I only had trouble with this one problem. I had to find the arclength of a polar equation with 0 < x < pi (x being substituted for theta for ease of typing). Anyway the original equation escapes me but I solved the problem to this point:

∫ √(4 - 3sin²x + (sinx)^4) dx

"the intregal from 0-pi of the square root of four minus 3 times sine squared of x plus sine of x raised to the 4th power "

I tried messing with sine and cosine Identities but after three tries of manipulating the numbers I kept coming back to this equation.

Where should I have gone from here?

I don't believe that integral has an elementary antiderivative. Are you sure the expression under the square root is correct? It looks like you could make the substitution u = sin(x)^2 and solve the quadratic, but u^2-3u+4 has no real roots.

You messed up somewhere as that integral is very difficult to evaluate. Try finding a similar example and i'm sure someone will go through it with you.

Ok so I just had a calculus final and I only had trouble with this one problem. I had to find the arclength of a polar equation with 0 < x < pi (x being substituted for theta for ease of typing). Anyway the original equation escapes me but I solved the problem to this point:

? ?(4 - 3sin²x + (sinx)^4) dx



sin^2x= 1/2(1-cos2x)
therefore your equation equals
(int.)4-3/2(1-cos2x)+(1/2(1-cos2x))^2

if that doesnt work use cos^2x=1/2(1+cos2x)

I took my calculus two final today ;)

Oh yeah, http://en.wikipedia.org/wiki/Trigonometric_identies#Power-reduction_formulas

you can reduce it to:

Integral(1/8 (23 + 8 Cos[2 x] + Cos[4 x])) dx

= (23 x)/8 + 1/2 Sin[2 x] + 1/32 Sin[4 x]

Sorry guys, but there's a square root sign there. There's no closed form solution to the integral. The function is analytic, so you could find the power series and integrate term by term. I.E. sqrt(4-3(sinx)^2+(sinx)^4 = 2 - 3/2x^2 + 69/8x^4 - ...
so int(sqrt(4-3(sinx)^2+(sinx)^4) = 2x - 1/2x^3 + 69/40x^5 - ....
and there might be methods for finding definite integrals, but this is just something you do with brute force on a computer, not on a calculus test.