∫ sin(x) cos(x) dx, ∫ cosh(x) / (cosh^2(x) - 1) dx, ∫ sinh(√x) / √x dx, ∫ cos(x) / √(1 - sin(x)) dx, ∫ x^3 / (√x + 1) dx, ∫ x Arctan(x) dx

Steps to Solve

1. Integral of sin(x) cos(x)

Use the identity $ \sin(2x) = 2\sin(x)\cos(x) $ to rewrite the integral. Then we have: $$ \int \sin(x)\cos(x) , dx = \frac{1}{2} \int \sin(2x) , dx $$ Integrate: $$ = -\frac{1}{4} \cos(2x) + C $$

2. Integral of cosh(x) / (cosh²(x) - 1)

Recognize that $ \cosh^2(x) - 1 = \sinh^2(x) $. Therefore: $$ \int \frac{\cosh(x)}{\sinh^2(x)} , dx = \int \csch^2(x) , dx $$ The integral of $\csch^2(x)$ is: $$ = -\coth(x) + C $$

3. Integral of sinh(√x) / √x

Use the substitution $ u = \sqrt{x} $, then $ du = \frac{1}{2\sqrt{x}} , dx $, or $ dx = 2u , du $. Thus: $$ \int \frac{\sinh(u)}{u} (2u) , du = 2 \int \sinh(u) , du $$ The result is: $$ = 2\cosh(u) + C = 2\cosh(\sqrt{x}) + C $$

4. Integral of cos(x) / √(1 - sin(x))

Use the substitution $ u = 1 - \sin(x) $, then $ -du = \cos(x) , dx $. Thus: $$ = -\int \frac{1}{\sqrt{u}} , du $$ This becomes: $$ = -2\sqrt{u} + C = -2\sqrt{1 - \sin(x)} + C $$

5. Integral of x³ / (√x + 1)

Rewrite the integral: $$ \int \frac{x^{3/2}}{\sqrt{x} + 1} , dx $$ by dividing: $$ = \int (x^{3/2} - x^{1/2}) , dx + \int \frac{x^{1/2}}{\sqrt{x} + 1} , dx $$ The first part integrates to: $$ \frac{2}{5}x^{5/2} - \frac{2}{3}x^{3/2} + C $$ The second requires substitution or expansion.

6. Integral of x Arctan(x)

Use integration by parts, where: $$ u = \arctan(x), \quad dv = x , dx $$ Then: $$ du = \frac{1}{1+x^2} , dx, \quad v = \frac{x^2}{2} $$ Applying integration by parts gives: $$ = \frac{x^2}{2} \arctan(x) - \int \frac{x^2/2}{1+x^2} , dx $$ Simplifying the remaining integral leads to a solution.