Prove that `int_0^(pi//2) sin 2x log tanx dx=0`

Answer

To prove that

\[ \int_0^{\frac{\pi}{2}} \sin 2x \log \tan x \, dx = 0, \]

we will use the property of integrals that states:

\[ \int_0^a f(x) \, dx = \int_0^a f(a - x) \, dx. \]

### Step 1: Define the Integral
Let

\[ I = \int_0^{\frac{\pi}{2}} \sin 2x \log \tan x \, dx. \]

### Step 2: Apply the Integral Property
Using the property of integrals, we can rewrite the integral as follows:

\[ I = \int_0^{\frac{\pi}{2}} \sin 2\left(\frac{\pi}{2} - x\right) \log \tan\left(\frac{\pi}{2} - x\right) \, dx. \]

### Step 3: Simplify the Integral
Now, we simplify the terms inside the integral:

- \(\sin 2\left(\frac{\pi}{2} - x\right) = \sin(\pi - 2x) = \sin 2x\) (since \(\sin\) is positive in the second quadrant).
- \(\tan\left(\frac{\pi}{2} - x\right) = \cot x\).

Thus, we can rewrite the integral:

\[ I = \int_0^{\frac{\pi}{2}} \sin 2x \log \cot x \, dx. \]

### Step 4: Rewrite \(\log \cot x\)
We know that:

\[ \log \cot x = \log \left(\frac{1}{\tan x}\right) = -\log \tan x. \]

### Step 5: Substitute Back into the Integral
Substituting this back, we have:

\[ I = \int_0^{\frac{\pi}{2}} \sin 2x (-\log \tan x) \, dx = -\int_0^{\frac{\pi}{2}} \sin 2x \log \tan x \, dx = -I. \]

### Step 6: Add the Two Expressions for \(I\)
Now, we add the two expressions for \(I\):

\[ I + I = 0 \implies 2I = 0 \implies I = 0. \]

### Conclusion
Thus, we have proved that:

\[ \int_0^{\frac{\pi}{2}} \sin 2x \log \tan x \, dx = 0. \]

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