When do we have to change the limits?∫01x-x2dxcompleting the square gives me:∫-x2+x-14+14dx∫-(x-12)2+14dx∫-u2+14du∫14-u2duIs trig sub the best way to go?if&nbsp;u=12sinθ&nbsp;then&nbsp;du=12cosθdθ=∫14-14sin2θ12cosθdθ∫12cosθ12cosθdθ14∫cos2θdθusing half angle identity I eventually get:14∫12(1+cos(2θ))dθ18∫(1+cos(2θ)dθ18(θ+12sin2θ)Where do I go from here?

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When do we have to change the limits?∫01x-x2dxcompleting the square gives me:∫-x2+x-14+14dx∫-(x-12)2+14dx∫-u2+14du∫14-u2duIs trig sub the best way to go?if&amp;nbsp;u=12sinθ&amp;nbsp;then&amp;nbsp;du=12cosθdθ=∫14-14sin2θ12cosθdθ∫12cosθ12cosθdθ14∫cos2θdθusing half angle identity I eventually get:14∫12(1+cos(2θ))dθ18∫(1+cos(2θ)dθ18(θ+12sin2θ)Where do I go from here?

Dakota Cunningham · Accepted Answer

I would recommend changing the limits every time we substitute. Here, that leads to ∫−1/21/214−u2du (u=x−12 runs from 0−12 to 1−12) 14∫−π/2π/2cos2⁡θdθ (u=12sin⁡θ so since sin⁡θ runs from −1 to 1, θ runs from −π2  to  π2) [18(θ+12sin⁡(2θ))]θ=−π2θ=π2=(π16+0)−(−π16+0)=π8 If you do it that way, there&#039;s no need to go back -- we just evaluate the integral right there. Definite integrals have more tools we can use than indefinite integrals. Even at the basic calculus level you&#039;re at, symmetry comes up a lot -- the integral of an odd function from −a to a is zero, for example. If we turn a definite integral problem into an indefinite integral problem and then only convert back at the very end, we lose access to those tools. So then, it makes sense to always bring the limits along and convert them when substituting in an indefinite integral problem. Is trig sub the best way to go? No, the best way to go is to not calculate any antiderivatives at all. Once you reach that first line I wrote down (the same point you were at when writing this line, plus the limits), recognize it as the area of a semicircle of radius 12 From that, we immediately conclude the result 12π(12)2=π8 That is, of course, not something we could ever have done with an indefinite integral.

mihady54 · Accepted Answer

Just so you are aware, this can be handled with the Beta and by extension the Gamma Function.   Here you have:  I=∫01x−x2:dx=∫01x(1−x):dx=∫01x12(1−x)12:dx   =B(12+1,12+1)=B(32,32)  Using the relationship between the Beta and the Gamma Function this becomes:  I=B(32,32)=Γ(32)⋅Γ(32)Γ(32+32)=Γ(32t)2Γ(3)=(π2)22=π8

int_0^1 sqrt(x-x^2) dx completing the square gives me: int sqrt(-x^2 +x - 1/4 + 1/4) dx, int sqrt(-(x-1/2)^2 + 1/4) dx

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