Mathematics: When i was 2 years old,my brother was half of my age..now im 100,my brother is?

1) 2) 3) Step-by-step explanation:1) This integral can be solved easily by using algebraic substitutions:, Then, the integral can rewritten as follows:Where is the integration constant.2) Let be , then we obtain the expression by the definition of the derivative for the inverse hyperbolic cosine and the chain rule:3) This integral can be solved by the following trigonometric substitutions:Where is the integration constant....

When i was 2 years old,my brother was half of my - id22362706, officialgraciela67

Ответ:

LuxTheGirl2543

22.05.2021

Mathematics

1) $\int\limits^{16}_{2} {\frac{dx}{2\cdot x \cdot \sqrt{\ln x}} } \approx 1.665$

2) $\frac{dy}{dx} = \pm \frac{1}{\sqrt{4\cdot x^{2}+7\cdot x +3}}$

3) $\int\limits^{2\sqrt{3}}_{0} {\frac{dx}{\sqrt{4 + x^{2}}} } \approx 1.317$

Step-by-step explanation:

1) $\int\limits^{16}_{2} {\frac{dx}{2\cdot x \cdot \sqrt{\ln x}} }$

This integral can be solved easily by using algebraic substitutions:

$u = \ln x$ , $du = \frac{dx}{x}$

Then, the integral can rewritten as follows:

$\int {\frac{dx}{2\cdot x\cdot \sqrt{\ln x}} } = \frac{1}{2}\int {\frac{du}{u^{1/2}} } = \frac{1}{2}\int {u^{-1/2}} \, du$

$\int {u^{-1/2}} \, du = 2\cdot u^{1/2} + C = 2\cdot \sqrt{\ln x} + C$

Where is the integration constant.

$\int\limits^{16}_{2} {\frac{dx}{2\cdot x \cdot \sqrt{\ln x}} } = F(16) - F(2)$

$F(16) - F(2) = 2\cdot (\sqrt{\ln 16}-\sqrt{\ln 2})$

$F(16) - F(2) \approx 1.665$

2) Let be $y = \cosh^{-1} (2\cdot \sqrt{x+1})$ , then we obtain the expression by the definition of the derivative for the inverse hyperbolic cosine and the chain rule:

$\frac{dy}{dx} = \pm\frac{1}{\sqrt{4\cdot x + 3}}\cdot \left(\frac{1}{\sqrt{x+1}} \right)$

$\frac{dy}{dx} = \pm \frac{1}{\sqrt{(4\cdot x + 3)\cdot (x+1)}}$

$\frac{dy}{dx} = \pm \frac{1}{\sqrt{4\cdot x^{2}+7\cdot x +3}}$

3) $\int\limits^{2\sqrt{3}}_{0} {\frac{dx}{\sqrt{4 + x^{2}}} }$

This integral can be solved by the following trigonometric substitutions:

$\frac{2}{\sqrt{4 + x^{2}}} = \cos \theta$

$\frac{1}{\sqrt{4+x^{2}}} = \frac{\cos \theta}{2}$

$\frac{x}{2} = \tan \theta$

$x = 2\cdot \tan \theta$

$dx = 2\cdot \sec^{2}\theta \,d \theta$

$\int {\frac{dx}{\sqrt{4+x^{2}}} } = \int {\left(\frac{\cos \theta}{2} \right)\cdot (2\cdot \sec^{2}\theta)} \, d\theta = \int {\sec \theta} \, d\theta$