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Chapter 7: Problem 28

Evaluate the following integrals. $$\int \frac{3 x+1}{\sqrt{4-x^{2}}} d x$$

Short Answer

Expert verified
Answer: The substitution used is \(x = 2\sin(u)\).

Step by step solution

01

Choosing the right substitution

Let's choose the substitution \(x = 2\sin(u)\). This substitution allows us to rewrite the expression in the denominator and simplify the integrand nicely. Now, we find the derivative of the substitution to replace \(dx\). So, \(dx = 2\cos(u) du\).
02

Apply the substitution to the integral

Now, we replace \(x\) with our substitution and \(dx\) with its corresponding expression: $$\int \frac{3x+1}{\sqrt{4-x^2}} dx = \int \frac{3(2\sin(u))+1}{\sqrt{4-(2\sin(u))^2}} (2\cos(u) du)$$
03

Simplify the integral

After substituting, the integral becomes: $$\int \frac{6\sin(u)+1}{\sqrt{4-(4\sin^2(u))}} (2\cos(u) du)$$ Now, let's simplify the expression inside the square root: $$\int \frac{6\sin(u)+1}{\sqrt{4(1-\sin^2(u))}} (2\cos(u) du)$$ Now, we can simplify further by using the identity \(\cos^2(u) = 1-\sin^2(u)\): $$\int \frac{6\sin(u)+1}{\sqrt{4\cos^2(u)}} (2\cos(u) du)$$ Finally we have, $$\int \frac{6\sin(u)+1}{2\cos(u)} (2\cos(u) du)$$
04

Cancel the common terms

We can cancel out the \(2\cos(u)\) in the numerator and denominator: $$\int (6\sin(u)+1) du$$
05

Evaluate the integral

Now, we need to find the antiderivative of the simplified integrand: $$\int (6\sin(u)+1) du = 6\int \sin(u) du + \int 1 du$$ Using the antiderivatives of the sine and constant functions, we obtain: $$= 6(-\cos(u)) + u + C$$
06

Replace the substitution

Now, we need to replace \(u\) back with our original substitution, \(x = 2\sin(u)\): $$= 6(-\cos(\sin^{-1}(x/2))) + \sin^{-1}(x/2) + C$$ So the integral evaluates to: $$\int \frac{3x+1}{\sqrt{4-x^2}} dx = 6(-\cos(\sin^{-1}(x/2))) + \sin^{-1}(x/2) + C$$

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Most popular questions from this chapter

A differential equation of the form \(y^{\prime}(t)=F(y)\) is said to be autonomous (the function \(F\) depends only on \(y\) ). The constant function \(y=y_{0}\) is an equilibrium solution of the equation provided \(F\left(y_{0}\right)=0\) (because then \(y^{\prime}(t)=0,\) and the solution remains constant for all \(t\) ). Note that equilibrium solutions correspond to horizontal line segments in the direction field. Note also that for autonomous equations, the direction field is independent of \(t\). Consider the following equations. a. Find all equilibrium solutions. b. Sketch the direction field on either side of the equilibrium solutions for \(t \geq 0\). c. Sketch the solution curve that corresponds to the initial condition \(y(0)=1\). $$y^{\prime}(t)=y(y-3)$$

Graph the integrands and then evaluate and compare the values of \(\int_{0}^{\infty} x e^{-x^{2}} d x\) and \(\int_{0}^{\infty} x^{2} e^{-x^{2}} d x\).

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Many methods needed Show that \(\int_{0}^{\infty} \frac{\sqrt{x} \ln x}{(1+x)^{2}} d x=\pi\) in the following steps. a. Integrate by parts with \(u=\sqrt{x} \ln x\) b. Change variables by letting \(y=1 / x\) c. Show that \(\int_{0}^{1} \frac{\ln x}{\sqrt{x}(1+x)} d x=-\int_{1}^{\infty} \frac{\ln x}{\sqrt{x}(1+x)} d x\) (and that both integrals converge). Conclude that \(\int_{0}^{\infty} \frac{\ln x}{\sqrt{x}(1+x)} d x-0\) d. Evaluate the remaining integral using the change of variables \(z=\sqrt{x}\)

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