Evaluate the iterated polar integral of 2cos(theta)sin(theta)/r^2 times r dr, with r from root(cos theta sin theta) to 1 and theta from 0 to pi/2.

Step-by-step solution

1. 1

   Simplify the integrand first. The polar area element already supplies the factor $r$, so it cancels one power in the denominator:

   $\frac{2\cos\theta\sin\theta}{r^{2}}\cdot r=\frac{2\cos\theta\sin\theta}{r}=\frac{\sin2\theta}{r}$

   using the double-angle identity $2\sin\theta\cos\theta=\sin2\theta$. A $\tfrac1r$ integrand is the signal that a logarithm is coming.

2. 2

   Do the inner integral in $r$. For fixed $\theta$ the numerator is constant:

   $\int_{\sqrt{\cos\theta\sin\theta}}^{1}\frac{\sin2\theta}{r}\,dr=\sin2\theta\Big[\ln r\Big]_{\sqrt{\cos\theta\sin\theta}}^{1}=-\sin2\theta\,\ln\sqrt{\cos\theta\sin\theta}$

   since $\ln1=0$. The lower limit is below $1$, so its logarithm is negative and the whole expression is positive.

3. 3

   Convert the logarithm using the double angle. Because $\cos\theta\sin\theta=\tfrac12\sin2\theta$,

   $-\ln\sqrt{\cos\theta\sin\theta}=-\tfrac12\ln\!\left(\tfrac12\sin2\theta\right)=-\tfrac12\left(\ln\sin2\theta-\ln2\right)$

   so the remaining integral is

   $I=-\frac12\int_{0}^{\pi/2}\sin2\theta\left(\ln\sin2\theta-\ln2\right)d\theta$

4. 4

   Substitute $u=2\theta$. With $d\theta=\tfrac{du}{2}$ and $u$ running from $0$ to $\pi$:

   $I=-\frac14\left[\int_{0}^{\pi}\sin u\,\ln\sin u\,du-\ln2\int_{0}^{\pi}\sin u\,du\right]$

5. 5

   Use the two standard integrals. The elementary one is $\displaystyle\int_{0}^{\pi}\sin u\,du=2$. The logarithmic one is the classical result

   $\int_{0}^{\pi}\sin u\,\ln\sin u\,du=2\ln2-2$

   Substituting both:

   $I=-\frac14\left[(2\ln2-2)-2\ln2\right]=-\frac14(-2)=\frac12$

6. 6

   Verify numerically. Simpson's rule applied to the reduced $\theta$-integrand $-\sin2\theta\ln\sqrt{\cos\theta\sin\theta}$ over $(0,\tfrac\pi2)$ gives $0.4999999990$ ✓, matching $\tfrac12$ to eight decimals. Note the integrand is integrable at both ends despite the logarithm blowing up there, because $\sin2\theta$ vanishes faster than $\ln$ diverges.