The technique known as Trigonometric Substitution is a powerful tool in the realm of calculus. It's a process that involves replacing a variable in an equation with a trigonometric function. The equation is then simplified, and the solution is found using inverse trigonometric functions. This method proves especially beneficial when dealing with integrals that have roots of quadratic expressions involved.
| Topic | Problem | Solution |
|---|---|---|
| None | Find $\frac{d y}{d r}$ for $y=\int_{0}^{r} \sqrt{… | Given the function \(y=\int_{0}^{r} \sqrt{11+12 t^{2}} d t\), we are asked to find \(\frac{d y}{d r… |
| None | Evaluate the integral. \[ \int \frac{e^{\sin ^{-1… | Let's start by substituting \(u = \sin^{-1}x\). This implies that \(du = \frac{dx}{\sqrt{1-x^{2}}}\… |
| None | Evaluate the integral $\int x^{3} \sqrt{x^{2}+7} … | Let's start by setting \(u = x^2 + 7\). |
| None | Evaluate the indefinite integral. \[ \int x^{3} \… | First, we recognize that this integral is in the form of a standard formula for integration by subs… |
| None | Evaluate the indefinite integral. \[ \int \frac{x… | Let's start by making a substitution. Let \(u = x^3\), then \(du = 3x^2 dx\). |
| None | Evaulate the integral by trigonometric substituti… | Let's use the trigonometric substitution method to solve this integral. The given integral is in th… |
| None | $\int \frac{2 \cos ^{4}(\sqrt{x}) \sin (\sqrt{x})… | Let $u = \sqrt{x}$, then $x = u^2$ and $dx = 2u du$ |
| None | $\int \frac{d z}{s^{2} \sqrt{s^{2}-9}}$ | Given the integral: \(\int \frac{d z}{s^{2} \sqrt{s^{2}-9}}\) |
| None | f) \( \int \sqrt[3]{3 x^{3}-5 x^{5}} d x \) | \( u(x)=3x^3-5x^5 \) |
| None | Question 9 (4 points) When evaluating the integra… | 1. Let \(u = 2x\) then \(\frac{du}{dx} = 2\) |
| None | Иант ํo 15 \[ \int \frac{3 x+5}{\sqrt{x^{2}+4 x+5… | Let \(u = x^2 + 4x + 5\), then \(\frac{du}{dx} = 2x + 4\) |