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I was calculating an integral that I thought it'd be easy however it turned out very weird. I couldn't figure out so I'm looking for some help.

$$\int_{-\pi/2}^{\pi/2} \int_{2\cos\theta}^{2}2r{\sqrt{4-r^2}} \,dr\,d\vartheta$$

Then, I solved the inner integral with respect to r. I got the following by simplifying

$\int {2\over3}(4-(2\cos\theta)^2)^{3\over2} $

$\int {2\over3}(4-4\cos^2\theta)^{3\over2} $

$\int {2\over3}(4(1-\cos^2\theta))^{3\over2} $

$\int {2\over3}(4\sin^2\theta)^{3\over2} $

${16\over3}\int(\sin^3\theta) $

Then, with simple substitution, I got $\int(\sin^3\theta) d\theta $ = $1/3\cos^3\theta-\cos\theta+C$. When I plug in my end points $\pi/2$ and $-\pi/2$, I get zero. However, it can't be zero because it's a legit solid piece. So, I tried to solve with wolphram alpha for $\sin^3\theta$ and my integrand before simplification, I got followings:

enter image description here

enter image description here

I also found this one for the integration of $\sin^3x$

enter image description here vs enter image description here

So, I'm super confused how come $\sin^3x$ got two different answers???They seem the same but how come they are completely different functions? Where do I make mistake to solve the original problem?

thank you

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  • $\begingroup$ $\frac { 1 }{ 12 } \left( \cos { 3x } -9\cos { x } \right) =-\frac { 9 }{ 12 } \cos { x } +\frac { 1 }{ 12 } \cos { 3x } =-\frac { 3 }{ 4 } \cos { x } +\frac { 1 }{ 12 } \cos { 3x } $ $\endgroup$
    – haqnatural
    Commented Jan 28, 2017 at 9:33

3 Answers 3

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There is no contradiction ...

If you use the substitution $y=\cos(x)$, you find that the primitives of $\sin^3$ are given by $x\mapsto\frac{1}{3}\cos^3(x)-\cos(x)+C$

On the other side, if you linearize first, you get $\sin^3(x)=\frac{1}{4}(3\sin(x)-\sin(3x))$, and hence the primitives of $\sin^3$ are given by $x\mapsto\frac{1}{12}\cos(3x)-\frac{3}{4}\cos(x)+C$

So, apparently, we get two different families of functions ... but only apparently because linearization of $\cos^3(x)$ leads to the formula :

$\cos^3(x)=\frac{1}{4}(\cos(3x)+3\cos(x))$

and so :

$\frac{1}{3}\cos^3(x)-\cos(x)=\frac{1}{12}(\cos(3x)+3\cos(x))-\cos(x)=\frac{1}{12}\cos(3x)-\frac{3}{4}\cos(x)$

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$\int\sin^3xdx=\int(1-\cos^2x)\sin xdx=\int(\cos^2x-1)d(\cos x)=\frac{\cos^3x}{3}-\cos x+C$

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There is actually no mistake. They are exactly the same since $\cos(3x)\equiv4\cos^3 x-3\cos x$. We demonstrate this as follows: $$\cos(3x)=\cos(2x+x)$$ Using the double angle formula: $$\cos(3x)=\cos(2x)\cdot\cos x-\sin(2x)\cdot \sin x$$ Using $\cos(2x)\equiv 2\cos^2 x-1$ and $\sin(2x)\equiv 2\sin x\cos x$, we obtain: $$\cos(3x)=2\cos^3 x-\cos x-2\sin^2 x\cos x$$ Using the identity $\sin^2 x\equiv 1-\cos^2 x$, we obtain the correct identity: $$\cos(3x)=2\cos^3 x-\cos x-2(1-\cos^2 x)\cos x$$ $$\boxed{\cos(3x)\equiv 4\cos^3 x-3\cos x}$$

Substituting this on your first solution gives the second solution.

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