Integral Calculation Exercise 12

We’re asked to find an antiderivative of the function

$$ F(x) = \int \frac{ \log^2[\sin(x)] }{\sin^2(x)} \cdot \sin(x) \ dt $$

with the condition

$$ F\left(\frac{\pi}{2}\right) = 1 $$

To enforce this condition, we set the lower limit of integration to $ \pi/2 $:

$$ F(x) = \int_{\frac{\pi}{2}}^x \frac{ \log^2[\sin(t)] }{\sin^2(t)} \cdot \sin(t) \ dt $$

Since this definite integral equals zero when $ x = \pi/2 $, we add 1 to satisfy the initial condition:

$$ F(x) = 1 + \int_{\frac{\pi}{2}}^x \frac{ \log^2[\sin(t)] }{\sin^2(t)} \cdot \sin(t) \ dt $$

This adjustment ensures that $ F(\pi/2) = 1 $ as required.

Let’s now compute the integral.

We apply the double-angle identity for sine: $\sin(2t) = 2\sin(t)\cos(t)$.

$$ F(x) = 1 + \int_{\frac{\pi}{2}}^x \frac{ \log^2[\sin(t)] }{\sin^2(t)} \cdot 2 \sin(t) \cos(t) \ dt $$

This allows us to cancel out a factor of $ \sin(t) $:

$$ F(x) = 1 + \int_{\frac{\pi}{2}}^x \frac{ \log^2[\sin(t)] }{\sin(t)} \cdot 2 \cos(t) \ dt $$

$$ F(x) = 1 + 2 \int_{\frac{\pi}{2}}^x \frac{ \log^2[\sin(t)] }{\sin(t)} \cos(t) \ dt $$

Notice that $ \frac{\cos(t)}{\sin(t)} \, dt $ is exactly the differential of $ \log[\sin(t)] $:

$$ F(x) = 1 + 2 \int_{\frac{\pi}{2}}^x \log^2[\sin(t)] \, d(\log[\sin(t)]) $$

Explanation. $$ \frac{d}{dt} \left[ \log(\sin(t)) \right] = \frac{1}{\sin(t)} \cdot \frac{d}{dt}[\sin(t)] = \frac{1}{\sin(t)} \cdot \cos(t) = \frac{\cos(t)}{\sin(t)} $$

Let $ s = \log(\sin(t)) $, so the integral becomes:

$$ F(x) = 1 + 2 \int_{\frac{\pi}{2}}^x s^2 \ ds $$

Now we evaluate the integral:

$$ F(x) = 1 + 2 \cdot \left[ \frac{s^3}{3} \right]_{\frac{\pi}{2}}^x $$

Substituting back $ s = \log(\sin(t)) $:

$$ F(x) = 1 + 2 \cdot \left[ \frac{ \log^3[\sin(t)] }{3} \right]_{\frac{\pi}{2}}^x $$

$$ F(x) = 1 + \frac{2}{3} \cdot \left[ \log^3[\sin(t)] \right]_{\frac{\pi}{2}}^x $$

Evaluating the difference:

$$ F(x) = 1 + \frac{2}{3} \cdot \left[ \log^3[\sin(x)] - \log^3\left[\sin\left(\frac{\pi}{2}\right)\right] \right] $$

Since $ \log^3[\sin(\pi/2)] = \log^3[1] = 0 $, we have:

$$ F(x) = 1 + \frac{2}{3} \cdot \log^3[\sin(x)] $$

And that concludes the solution.