`int_(0)^(pi/6) sin2x . cosx dx `

Answer

To evaluate the integral \( I = \int_{0}^{\frac{\pi}{6}} \sin(2x) \cos(x) \, dx \), we can follow these steps:

### Step 1: Rewrite the Integral
Using the double angle identity for sine, we can express \( \sin(2x) \) as \( 2 \sin(x) \cos(x) \). Thus, we rewrite the integral:
\[ I = \int_{0}^{\frac{\pi}{6}} \sin(2x) \cos(x) \, dx = \int_{0}^{\frac{\pi}{6}} 2 \sin(x) \cos(x) \cos(x) \, dx = 2 \int_{0}^{\frac{\pi}{6}} \sin(x) \cos^2(x) \, dx \]

### Step 2: Substitution
Let \( t = \cos(x) \). Then, the derivative \( dt = -\sin(x) \, dx \), which implies \( \sin(x) \, dx = -dt \). We also need to change the limits of integration:
- When \( x = 0 \), \( t = \cos(0) = 1 \)
- When \( x = \frac{\pi}{6} \), \( t = \cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2} \)

Now, substituting these into the integral gives:
\[ I = 2 \int_{1}^{\frac{\sqrt{3}}{2}} \sin(x) \cos^2(x) \, dx = 2 \int_{1}^{\frac{\sqrt{3}}{2}} \cos^2(x) (-dt) \]
Reversing the limits of integration introduces a negative sign:
\[ I = 2 \int_{\frac{\sqrt{3}}{2}}^{1} t^2 (-dt) = 2 \int_{\frac{\sqrt{3}}{2}}^{1} t^2 \, dt \]

### Step 3: Evaluate the Integral
Now we can compute the integral:
\[ I = 2 \int_{\frac{\sqrt{3}}{2}}^{1} t^2 \, dt = 2 \left[ \frac{t^3}{3} \right]_{\frac{\sqrt{3}}{2}}^{1} \]
Calculating this gives:
\[ = 2 \left( \frac{1^3}{3} - \frac{\left(\frac{\sqrt{3}}{2}\right)^3}{3} \right) = 2 \left( \frac{1}{3} - \frac{3\sqrt{3}}{24} \right) = 2 \left( \frac{1}{3} - \frac{\sqrt{3}}{8} \right) \]

### Step 4: Simplify the Expression
To simplify:
\[ = 2 \left( \frac{8}{24} - \frac{3\sqrt{3}}{24} \right) = 2 \cdot \frac{8 - 3\sqrt{3}}{24} = \frac{16 - 6\sqrt{3}}{24} = \frac{8 - 3\sqrt{3}}{12} \]

Thus, the final answer is:
\[ I = \frac{8 - 3\sqrt{3}}{12} \]

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