Question

∫dx/√3-9x^2 решите подробно, пожалуйста! Очень срочно!!!

Answer 1

$\int\limits { \frac{1}{ \sqrt{3-9x^2} } } \, dx$
Очень похоже на арксинус, но надо немного довести выражение, чтобы интеграл стал арксинусом.

$\int\limits { \frac{1}{ \sqrt{3-9x^2} } } \, dx = \frac{1}{ \sqrt{3} } \int\limits { \frac{1}{ \sqrt{1-3x^2} } } \, dx = \frac{1}{ \sqrt{3} } \int\limits { \frac{1}{ \sqrt{1-( \sqrt{3} x)^2} } } \, dx = \\ \\ \frac{1}{ \sqrt{3} } \int\limits { \frac{ \frac{1}{\sqrt{3}} }{ \sqrt{1-( \sqrt{3} x)^2} } } \, d(\sqrt{3}x) = \frac{1}{3} \int\limits { \frac{ 1 }{ \sqrt{1-( \sqrt{3} x)^2} } } \, d(\sqrt{3}x)$

Вот теперь выражение приведено к тому, что интеграл будет равен арксинусу:

$\frac{1}{3} \int\limits { \frac{ 1 }{ \sqrt{1-( \sqrt{3} x)^2} } } \, d(\sqrt{3}x) = \frac{1}{3} arcsin\sqrt{3} x+C$

Answer 2

Solving the Integral ∫dx/√(3-9x^2)

To solve the given integral ∫dx/√(3-9x^2), we can use the trigonometric substitution method. Let's start by expressing the integral in terms of trigonometric functions.

We can rewrite the integral as: ∫dx/√(3-9x^2) = ∫dx/√(3(1-3x^2))

Now, we can use the substitution: x = (1/√3)sin(θ) dx = (1/√3)cos(θ)dθ

Substituting x and dx, the integral becomes: ∫(1/√3)cos(θ)dθ/√(3(1-3((1/√3)sin(θ))^2)) = (1/3)∫cos(θ)dθ/√(1-sin^2(θ)) = (1/3)∫cos(θ)dθ/√(cos^2(θ)) = (1/3)∫dθ

Integrating with respect to θ, we get: (1/3)θ + C

Now, we need to express the result in terms of x. Since x = (1/√3)sin(θ), we have sin(θ) = √3x. Using this, we can find θ in terms of x.

sin(θ) = √3x θ = arcsin(√3x)

Therefore, the final result is: (1/3)arcsin(√3x) + C

So, the solution to the integral ∫dx/√(3-9x^2) is (1/3)arcsin(√3x) + C.