Using Tables to Find Antiderivatives

Practice 1: The integral $\int \frac{1}{25-x^{2}} \mathrm{dx}$ matches table entry 37: $\int \frac{1}{a^{2}-x^{2}} d x=\frac{1}{2 a} \ln \left|\frac{a+x}{a-x}\right|+C$

when we put $a=5$, so $\int \frac{1}{25-x^{2}} \mathrm{dx}=\frac{1}{2 \cdot 5} \ln \left|\frac{5+\mathrm{x}}{5-\mathrm{x}}\right|+\mathrm{C}$.

Practice 2: The integral $\int \frac{1}{7-x^{2}} \mathrm{dx}$ matches table entry 37: $\int \frac{1}{a^{2}-x^{2}} d x=\frac{1}{2 a} \ln \left|\frac{a+x}{a-x}\right|+C$

when $\mathrm{a}=\sqrt{7}, \text { so } \int \frac{1}{7-x^{2}} \mathrm{dx}=\frac{1}{2 \sqrt{7}} \ln \left|\frac{\sqrt{7}+\mathrm{x}}{\sqrt{7-x}}\right|+\mathrm{C}$.

Practice 3: $\int \frac{1}{25-9 x^{2}} \mathrm{dx}=\frac{1}{9} \int \frac{1}{(25 / 9)-x^{2}} \mathrm{dx}=\frac{1}{9} \int \frac{1}{(5 / 3)^{2}-x^{2}} \mathrm{dx}$

so use #37 with $a = 5/3$. Then

$\int \frac{1}{25-9 x^{2}} \mathrm{dx}=\frac{1}{9} \int \frac{1}{(5 / 3)^{2}-x^{2}} \mathrm{dx}=\frac{1}{9} \frac{1}{2(5 / 3)} \ln \left|\frac{5 / 3+\mathrm{x}}{5 / 3-\mathrm{x}}\right|+\mathrm{C}$

$=\frac{1}{30} \ln \left|\frac{5 / 3+\mathrm{x}}{5 / 3-\mathrm{x}}\right|+\mathrm{C}$  or $\frac{1}{30} \ln \left|\frac{5+3 \mathrm{x}}{5-3 \mathrm{x}}\right|+\mathrm{C}$

Practice 4: $\int \frac{\cos (x)}{25-\sin ^{2}(x)} \mathrm{dx}$. Put $\mathbf{u}=\sin (x)$. Then $\mathrm{du}=\cos (\mathrm{x}) \mathrm{d} \mathrm{x}$ so

$\int \frac{\cos (x)}{25-\sin ^{2}(x)} \mathrm{dx}=\int \frac{1}{25-\mathrm{u}^{2}} \mathrm{du}=\frac{1}{2 \cdot 5} \ln \left|\frac{5+\mathrm{u}}{5-\mathrm{u}}\right|+\mathrm{C}=\frac{1}{10} \ln \left|\frac{5+\sin (\mathrm{x})}{5-\sin (\mathrm{x})}\right|+\mathrm{C}$

Practice 5: We can evaluate $\int \cos ^{3}(7 x) d x$ by using the recursion formula from the Table with $n = 3$:

$\text { 20. } \int \cos ^{n}(a x) d x=\frac{\cos ^{n-1}(a x) \sin (a x)}{n a}+\frac{n-1}{n} \int \cos ^{n-2}(a x) d x \text {. }$ Then

$\begin{aligned} \int \cos ^{3}(7 x) \mathrm{dx}=& \frac{\cos ^{2}(7 \mathrm{x}) \cdot \sin (7 \mathrm{x})}{7 \cdot 3}+\frac{2}{3} \int \cos (7 x) \mathrm{d} x \\ &=\frac{\cos ^{2}(7 x) \cdot \sin (7 x)}{21}+\frac{2}{3} \cdot \frac{1}{7} \cdot \sin (7 x)+\mathrm{C} \end{aligned}$