Activity 63.
Evaluate each of the following indefinite integrals.
(a)
\(\displaystyle \int \sin(8-3x) \, dx\)
Hint.
Think \(\int \sin(u) \, du\text{.}\)
Answer.
\(\displaystyle \int \sin(8-3x) \, dx = -\frac{1}{3} (-\cos(8-3x)) + C\text{.}\)
Solution.
Since \(u=8-3x\) is linear and \(\int \sin(u) \, du = -\cos(u) + C\text{,}\) it follows that \(\int \sin(8-3x) \, dx = -\frac{1}{3} (-\cos(8-3x)) + C\text{.}\)
(b)
\(\displaystyle \int \sec^2 (4x) \, dx\)
Hint.
Think \(\int \sec^2 (u) \, du\text{.}\)
Answer.
\(\displaystyle \int \sec^2 (4x) \, dx = \frac{1}{4} \tan(4x) + C\text{.}\)
Solution.
Since \(4x\) is linear and \(\int \sec^2(u) \, du = \tan(u) + C \text{,}\) we see \(\int \sec^2 (4x) \, dx = \frac{1}{4} \tan(4x) + C\text{.}\)
(c)
\(\displaystyle \int \frac{1}{11x - 9} \, dx\)
Hint.
Think \(\int \frac{1}{u}\, du\text{.}\)
Answer.
\(\displaystyle \int \frac{1}{11x - 9} \, dx = \frac{1}{11} \ln|11x - 9| + C\text{.}\)
Solution.
Using the fact that \(\int \frac{1}{u} \, du = \ln |u| + C\text{,}\) we have \(\int \frac{1}{11x - 9} \, dx = \frac{1}{11} \ln|11x - 9| + C\text{.}\)
(d)
\(\displaystyle \int \csc(2x+1) \cot(2x+1) \, dx\)
Hint.
Think \(\int \csc(u) \cot(u) \, du\text{.}\)
Answer.
\(\displaystyle \int \csc(2x+1) \cot(2x+1) \, dx = -\frac{1}{2}\csc(2x+1) + C\text{.}\)
Solution.
We know \(\int \csc(u) \cot(u) \, du = -\csc(u) + C\text{,}\) so \(\int \csc(2x+1) \cot(2x+1) \, dx = -\frac{1}{2}\csc(2x+1) + C\text{.}\)
(e)
\(\displaystyle \int \frac{1}{\sqrt{1-16x^2}}\, dx\)
Hint.
Think \(\int \frac{1}{\sqrt{1-u^2}} \, du\text{.}\)
Answer.
\(\displaystyle \int \frac{1}{\sqrt{1-16x^2}}\, dx = \frac{1}{4} \arcsin(4x) + C\)
Solution.
Observe that \(\int \frac{1}{\sqrt{1-u^2}}\, dx = \arcsin(u) + C\text{,}\) and thus viewing \(16x^2 = (4x)^2\text{,}\) we see that \(\int \frac{1}{\sqrt{1-16x^2}}\, dx = \frac{1}{4} \arcsin(4x) + C\)
(f)
\(\displaystyle \int 5^{-x}\, dx\)
Hint.
Think \(\int 5^{u} \, du\text{.}\)
Answer.
\(\displaystyle \int 5^{-x}\, dx = -\frac{1}{\ln(5)}5^{-x} + C\text{.}\)
Solution.
Since \(\int 5^{u}\, du = \frac{1}{\ln(5)}5^u + C\text{,}\) we have that \(\int 5^{-x}\, dx = -\frac{1}{\ln(5)}5^{-x} + C\text{.}\)
