Problem 28 - Calculus Early Transcendentals

Question

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

Step-by-Step Solution

Answer

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

Step 1: 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, \(dx = 2\cos(u) du\).

Step 2: Apply the substitution to the integral

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

Step 3: 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, $$\int \frac{6\sin(u)+1}{2\cos(u)} (2\cos(u) du)$$

Step 4: Cancel the common terms

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

Step 5: 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: $$= 6(-\cos(u)) + u + C$$

Step 6: 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: $$\int \frac{3x+1}{\sqrt{4-x^2}} dx = 6(-\cos(\sin^{-1}(x/2))) + \sin^{-1}(x/2) + C$$