Problem 27 - Calculus Early Transcendentals

Question

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

Step-by-Step Solution

Answer

Question: Evaluate the integral $\int \frac{2 - 3 x}{\sqrt{1-x^2}} \, d x$. Answer: $2 \sin^{-1}(x) + 3\sqrt{1 - x^2} + C$

Step 1: Choose the substitution

We use the substitution \(x = \sin u\). Then, we find the differential \(dx\) in terms of \(du\): $$d x = \cos u \, d u$$

Step 2: Rewrite the integral

Replace \(x\) with \(\sin u\) and \(dx\) with \(\cos u \, du\) in the given integral: $$\int \frac{2 - 3 \sin u}{\sqrt{1-\sin^2 u}} \cos u \, d u$$

Step 3: Simplify the integrand

Notice that the denominator simplifies since \(1 - \sin^2 u = \cos^2 u\) and taking the square root of this expression gives us \(\cos u\): $$\int \frac{2 - 3 \sin u}{\cos u} \cos u \, d u$$ Now, the \(\cos u\) in the denominator and numerator cancel out, leaving us with the integrand: $$\int (2 - 3 \sin u) \, d u$$

Step 4: Integrate with respect to \(u\)

Now we can integrate term by term: $$\int (2 - 3 \sin u) \, d u = 2 \int 1 \, d u - 3 \int \sin u \, d u$$ The integral of \(1\) with respect to \(u\) is \(u\) and the integral of \(\sin u\) with respect to \(u\) is \(-\cos u\): $$2u + 3\cos u + C$$

Step 5: Replace \(u\) with \(x\)

Recall that we substituted \(x = \sin u\), so we have \(u=\sin^{-1}(x)\). Now replace \(u\) with \(\sin^{-1}(x)\): $$2\sin^{-1}(x) + 3\cos(\sin^{-1}(x)) + C$$

Step 6: Simplify the expression

Using the Pythagorean trigonometric identity, we can rewrite \(\cos(\sin^{-1}(x))\) in terms of \(x\). Recall that \(\cos^2 t = 1 - \sin^2 t\) for any angle \(t\). Hence, $$\cos(\sin^{-1}(x)) = \sqrt{1 - x^2}$$ Now plug this back into the expression we got in the previous step: $$2 \sin^{-1}(x) + 3\sqrt{1 - x^2} + C$$ This is our final answer. Thus, the integral of the given function is: $$\int \frac{2 - 3 x}{\sqrt{1-x^2}} \, d x = 2 \sin^{-1}(x) + 3\sqrt{1 - x^2} + C$$