Question

1. use guess and fudge to find the antiderivatives:

(i) $\int \cos (3 \pi x) d x$
(v) $\int 4 x^{2}\left(x^{3}+3\
ight)^{7} d x$
(ii) $\int(3 x+1)^{5} d x$
(vi) $\int x \sec ^{2}\left(x^{2}+1\
ight) d x$
(iii) $\int \frac{1}{(3 x+5)^{2}} d x$
(vii) $\int \frac{2 x}{\left(x^{2}+1\
ight)} d x$
(iv) $\int 2 x e^{-x^{2}} d x$

Explanation:

Step1: Antiderivative of \(\cos(3\pi x)\)

We know that the antiderivative of \(\cos(u)\) is \(\sin(u)\). Let \(u = 3\pi x\), then \(du=3\pi dx\), \(dx=\frac{du}{3\pi}\).
\(\int\cos(3\pi x)dx=\frac{1}{3\pi}\int\cos(u)du\)
Since \(\int\cos(u)du=\sin(u)+C\), substituting back \(u = 3\pi x\) gives \(\frac{1}{3\pi}\sin(3\pi x)+C\)

Step2: Antiderivative of \((3x + 1)^{5}\)

Let \(u=3x + 1\), then \(du = 3dx\), \(dx=\frac{du}{3}\)
\(\int(3x + 1)^{5}dx=\frac{1}{3}\int u^{5}du\)
Using the power - rule \(\int u^{n}du=\frac{u^{n + 1}}{n+1}+C\) (\(n
eq- 1\)), for \(n = 5\), \(\int u^{5}du=\frac{u^{6}}{6}+C\)
Substituting back \(u = 3x + 1\) gives \(\frac{(3x + 1)^{6}}{18}+C\)

Step3: Antiderivative of \(\frac{1}{(3x + 5)^{2}}\)

Let \(u=3x + 5\), then \(du=3dx\), \(dx=\frac{du}{3}\)
\(\int\frac{1}{(3x + 5)^{2}}dx=\frac{1}{3}\int u^{-2}du\)
Using the power - rule \(\int u^{n}du=\frac{u^{n + 1}}{n+1}+C\) (\(n
eq-1\)), for \(n=-2\), \(\int u^{-2}du=-\frac{1}{u}+C\)
Substituting back \(u = 3x + 5\) gives \(-\frac{1}{3(3x + 5)}+C\)

Step4: Antiderivative of \(2xe^{-x^{2}}\)

Let \(u=-x^{2}\), then \(du=-2xdx\), \(2xdx=-du\)
\(\int2xe^{-x^{2}}dx=-\int e^{u}du\)
Since \(\int e^{u}du=e^{u}+C\), substituting back \(u=-x^{2}\) gives \(-e^{-x^{2}}+C\)

Step5: Antiderivative of \(4x^{2}(x^{3}+3)^{7}\)

Let \(u=x^{3}+3\), then \(du = 3x^{2}dx\), \(x^{2}dx=\frac{du}{3}\)
\(\int4x^{2}(x^{3}+3)^{7}dx=\frac{4}{3}\int u^{7}du\)
Using the power - rule \(\int u^{n}du=\frac{u^{n + 1}}{n + 1}+C\) (\(n
eq-1\)), for \(n = 7\), \(\int u^{7}du=\frac{u^{8}}{8}+C\)
Substituting back \(u=x^{3}+3\) gives \(\frac{(x^{3}+3)^{8}}{6}+C\)

Step6: Antiderivative of \(x\sec^{2}(x^{2}+1)\)

Let \(u=x^{2}+1\), then \(du = 2xdx\), \(xdx=\frac{du}{2}\)
\(\int x\sec^{2}(x^{2}+1)dx=\frac{1}{2}\int\sec^{2}(u)du\)
Since \(\int\sec^{2}(u)du=\tan(u)+C\), substituting back \(u=x^{2}+1\) gives \(\frac{1}{2}\tan(x^{2}+1)+C\)

Step7: Antiderivative of \(\frac{2x}{x^{2}+1}\)

Let \(u=x^{2}+1\), then \(du = 2xdx\)
\(\int\frac{2x}{x^{2}+1}dx=\int\frac{du}{u}\)
Since \(\int\frac{1}{u}du=\ln|u|+C\), substituting back \(u=x^{2}+1\) gives \(\ln(x^{2}+1)+C\)

Answer:

(i) \(\frac{1}{3\pi}\sin(3\pi x)+C\)

(ii) \(\frac{(3x + 1)^{6}}{18}+C\)

(iii) \(-\frac{1}{3(3x + 5)}+C\)

(iv) \(-e^{-x^{2}}+C\)

(v) \(\frac{(x^{3}+3)^{8}}{6}+C\)

(vi) \(\frac{1}{2}\tan(x^{2}+1)+C\)

(vii) \(\ln(x^{2}+1)+C\)