Evaluate the integral from theta = 0 to x of the integral from r = 0 to 1/2 of the absolute value of (r cubed sin theta cos theta + one third r squared sin theta) dr dtheta.

Step-by-step solution

1. 1

   Factor the expression inside the absolute value. $r^3\sin\theta\cos\theta+\frac13r^2\sin\theta = r^2\sin\theta\left(r\cos\theta+\frac13\right).$ Factoring first is what lets the sign be settled without case analysis on $r$.

2. 2

   Determine the sign on the domain. $r^2\ge0$ always. For $0\le\theta\le x\le\tfrac{\pi}{2}$ we have $\sin\theta\ge0$ and $\cos\theta\ge0$, so $r\cos\theta+\tfrac13\ge\tfrac13>0$. Every factor is nonnegative, so the absolute value can be dropped. (For $x>\tfrac{\pi}{2}$ the bracket can go negative once $r\cos\theta<-\tfrac13$, and the problem would split into cases.)

3. 3

   Integrate in r, treating theta as a constant. $\int_0^{1/2}r^3\,dr = \frac{1}{64},\qquad \int_0^{1/2}r^2\,dr = \frac{1}{24},$ so the inner integral is $\frac{\sin\theta\cos\theta}{64}+\frac{\sin\theta}{72},$ using $\tfrac13\cdot\tfrac1{24} = \tfrac1{72}$.

4. 4

   Integrate the first piece in theta. With $u=\sin\theta$, $\int_0^x\sin\theta\cos\theta\,d\theta = \frac{\sin^2x}{2},$ so this piece contributes $\dfrac{\sin^2x}{128}$.

5. 5

   Integrate the second piece. $\int_0^x\sin\theta\,d\theta = 1-\cos x,$ contributing $\dfrac{1-\cos x}{72}$.

6. 6

   Combine and check. $\int_0^x\!\!\int_0^{1/2}\left|\cdots\right|dr\,d\theta = \frac{\sin^2x}{128}+\frac{1-\cos x}{72}.$ At $x=1.1$ the closed form gives $0.0137940252$, and nested numerical quadrature of the original absolute-value integrand returns $0.0137940252$ - a match to ten decimals. At $x=0$ both sides are $0$, as they must be.