1) ∫ cos²x dx = ? 0 to π/2 2) ∫ du/(e^x + e^−x) = ? 3) ∫ (tan⁻¹x)²/(1 - x²) dx = ? 0 to 1 4) ∫ sin¹⁷x * cos³x dx = ? -1 to 1 5) ∫ sinx⁰ dx = ? π/2 6) ∫ log(tanx) dx = ? 0 to... 1) ∫ cos²x dx = ? 0 to π/2 2) ∫ du/(e^x + e^−x) = ? 3) ∫ (tan⁻¹x)²/(1 - x²) dx = ? 0 to 1 4) ∫ sin¹⁷x * cos³x dx = ? -1 to 1 5) ∫ sinx⁰ dx = ? π/2 6) ∫ log(tanx) dx = ? 0 to π/2 7) ∫ secx dx/(secx + tanx) = ? 8) ∫ dx/(1 - sinx) = ? 9) ∫ cos³x * sinx dx = ? 10) ∫ logx dx = ? Read more

Steps to Solve

1. Setup the Integral We want to evaluate the integral

$$ \int_0^{\frac{\pi}{2}} \cos^2 x , dx $$

2. Use Trigonometric Identity We can use the identity

$$ \cos^2 x = \frac{1 + \cos(2x)}{2} $$

to rewrite the integral:

$$ \int_0^{\frac{\pi}{2}} \cos^2 x , dx = \int_0^{\frac{\pi}{2}} \frac{1 + \cos(2x)}{2} , dx $$

3. Separate the Integral This allows us to separate the integral:

$$ \int_0^{\frac{\pi}{2}} \frac{1}{2} , dx + \int_0^{\frac{\pi}{2}} \frac{\cos(2x)}{2} , dx $$

4. Evaluate Each Integral Now we evaluate each integral separately.

• For the first integral:

$$ \int_0^{\frac{\pi}{2}} \frac{1}{2} , dx = \frac{1}{2} \cdot \left[\frac{\pi}{2} - 0\right] = \frac{\pi}{4} $$

• For the second integral, we use the integration of cosine:

The integral of $\cos(2x)$ is $\frac{1}{2} \sin(2x)$, thus:

$$ \int_0^{\frac{\pi}{2}} \frac{\cos(2x)}{2} , dx = \frac{1}{2} \cdot \left[ \frac{1}{2} \sin(2x) \right]_0^{\frac{\pi}{2}} = \frac{1}{4} \left( \sin(\pi) - \sin(0) \right) = 0 $$

5. Combine the Results Adding both results together gives:

$$ \frac{\pi}{4} + 0 = \frac{\pi}{4} $$