1) ∫ cos² x dx = ? 2) ∫ du/(e^x + e^(-x)) = ? 3) ∫ (tan⁻¹ x)²/(1 - x²) dx = ? 4) ∫ sin¹⁷ x cos³ x dx = ? 5) ∫ sinⁿ x dx = ? 6) ∫[0 to π/2] log tan x dx = ? 7) ∫ sec x dx/(sec... 1) ∫ cos² x dx = ? 2) ∫ du/(e^x + e^(-x)) = ? 3) ∫ (tan⁻¹ x)²/(1 - x²) dx = ? 4) ∫ sin¹⁷ x cos³ x dx = ? 5) ∫ sinⁿ x dx = ? 6) ∫[0 to π/2] log tan x dx = ? 7) ∫ sec x dx/(sec x + tan x) = ? 8) ∫ dx/(1 - sin x) = ? 9) ∫ cos³ x sin x dx = ? 10) ∫ log x dx = ? Read more

Steps to Solve

1. First Integral: ∫ cos² x dx

Use the identity:
$$ \cos^2 x = \frac{1 + \cos(2x)}{2} $$

Then, integrate:
$$ \int \cos^2 x , dx = \int \frac{1 + \cos(2x)}{2} , dx = \frac{1}{2} \int 1 , dx + \frac{1}{2} \int \cos(2x) , dx $$
This results in:
$$ = \frac{x}{2} + \frac{1}{4} \sin(2x) + C $$

2. Second Integral: ∫ du/(e^u + e^(-u))

Rewrite the integral:
$$ \int \frac{du}{e^u + e^{-u}} = \int \frac{du}{2 \cosh(u)} $$

Then, it simplifies to:
$$ = \frac{1}{2} \int \text{sech}(u) , du = \frac{1}{2} \ln|\tanh(\frac{u}{2})| + C $$

3. Third Integral: ∫ (tan⁻¹ x)²/(1 - x²) dx

Use the substitution $u = \tan^{-1}(x)$, then $dx = \frac{1}{1+x^2} , du$:
$$ = \int u^2 \frac{du}{1-\tan^2(u)} $$

This is a complicated integration, requiring integration by parts.

4. Fourth Integral: ∫ sin¹⁷ x cos³ x dx

Use the substitution:
$$ u = \sin(x), , du = \cos(x) , dx $$
Then the integral becomes:
$$ \int u^{17} (1 - u^2)^{3/2} , du $$

5. Fifth Integral: ∫ sinⁿ x dx

Use reduction formulas, depending on whether $n$ is even or odd. If $n$ is even, relate to $\cos(x)$. If odd, factor out $\cos(x)$.

6. Sixth Integral: ∫[0 to π/2] log tan x dx

Use the property of definite integrals:
$$ I = \int_0^{\frac{\pi}{2}} \log(\tan x) , dx = -\frac{\pi}{2} \log(2) $$

7. Seventh Integral: ∫ sec x dx/(sec x + tan x)

Using substitution:
$$ u = sec(x) + tan(x), , du = (sec(x)tan(x) + sec^2(x)) dx $$
This transforms into:
$$ \int \frac{1}{u} , du = \ln|sec(x) + tan(x)| + C $$

8. Eighth Integral: ∫ dx/(1 - sin x)

Using the identity:
$$ 1 - \sin x = \frac{(1 - \sin x)^2}{1 + \sin x} $$
This results in a transformation and taking elementary parts.

9. Ninth Integral: ∫ cos³ x sin x dx

Use substitution:
$$ u = \cos(x), , du = -\sin(x) , dx $$ Integrate:
$$ -\int u^3 , du = -\frac{u^4}{4} + C = -\frac{\cos^4 x}{4} + C $$

10. Tenth Integral: ∫ log x dx

Use integration by parts:
Let $u = \log x \Rightarrow du = \frac{1}{x} , dx$ and $dv = dx$ gives:
$$ = x \log x - \int \frac{1}{x} , dx $$
This simplifies to:
$$ = x \log x - x + C $$