Chapter 7, Problem 1CRE

To determine

To match the integrals (a)-(e) with their anti-derivatives (i)-(v) on the basis of the general form (do no evaluate the integrals).

Answer to Problem 1CRE

Solution:

We have hence determined that (a) matches (v), (b) matches (iv), (c) matches (iii), (d) matches (i) and (e) matches (ii)

Explanation of Solution

Compare the integrals and the functions without evaluating the integrals to identify the correct match

Given:

(a) ${\displaystyle \int \frac{xdx}{x^2-4}}$ (b) ${\displaystyle \int \frac{(2x+9)dx}{x^2+4}}$ (c) ${\displaystyle \int {\sin }^{3}x{\cos }^{2}xdx}$ (d) ${\displaystyle \int \frac{dx}{x\sqrt{16x^2-1}}}$

(e) ${\displaystyle \int \frac{16dx}{x{(x-4)}^2}}$

(i) ${\sec }^{-1}4x+C$ (ii) $\log \left|x\right|-\log \left|x-4\right|-\frac{4}{x-4}+C$

(iii) $\frac{1}{30}(3{\cos }^{5}x-3{\cos }^{3}x\sin {}^{2}x-7{\cos }^{3}x)+C$

(iv) $\frac{9}{2}{\tan }^{-1}\frac{x}{2}+\ln (x^2+4)+C$ (v) $\sqrt{x^2-4}+C$

Calculation:

(a) ${\displaystyle \int \frac{xdx}{x^2-4}}$ - since $x$ is a constant multiple of the derivative $x^2-4$, the substitution method implies that the integral is a constant multiple of ${\displaystyle \int \frac{du}{\sqrt{u}},u=x^2-4}$ that is a constant multiple of $\sqrt{u}$. This hence matches the function in (v)

(b) ${\displaystyle \int \frac{(2x+9)dx}{x^2+4}}$ - ${\displaystyle \int \frac{2xdx}{x^2+4}}$corresponds to $\ln (x^2+4)$ and hence matches (iv)

(c) ${\displaystyle \int {\sin }^{3}x{\cos }^{2}xdx}$ - the reduction formula shows that this integral is the sum of constant multiples of products and hence matches the function in (iii)

(d) ${\displaystyle \int \frac{dx}{x\sqrt{16x^2-1}}}$- since ${\displaystyle \int \frac{dx}{\left|x\right|\sqrt{x^2-1}}}={\sec }^{-1}x+C$which corresponds to the function in (i)

(e) ${\displaystyle \int \frac{16dx}{x{(x-4)}^2}}$- This one matches the function in (ii)

Conclusion:

We have hence determined that (a) matches (v), (b) matches (iv), (c) matches (iii), (d) matches (i) and (e) matches (ii).