8. use guess and fudge to find the antiderivatives:\n(i) $\\int \\cos (3 \\pi x) d x$\n(v) $\\int 4…

8. use guess and fudge to find the antiderivatives:\n(i) $\\int \\cos (3 \\pi x) d x$\n(v) $\\int 4 x^{2}\\left(x^{3}+3\\right)^{7} d x$\n(ii) $\\int(3 x+1)^{5} d x$\n(vi) $\\int x \\sec ^{2}\\left(x^{2}+1\\right) d x$\n(iii) $\\int \\frac{1}{(3 x+5)^{2}} d x$\n(vii) $\\int \\frac{2 x}{\\left(x^{2}+1\\right)} d x$\n(iv) $\\int 2 x e^{-x^{2}} d x$\n

8. use guess and fudge to find the antiderivatives:\n(i) $\\int \\cos (3 \\pi x) d x$\n(v) $\\int 4 x^{2}\\left(x^{3}+3\\right)^{7} d x$\n(ii) $\\int(3 x+1)^{5} d x$\n(vi) $\\int x \\sec ^{2}\\left(x^{2}+1\\right) d x$\n(iii) $\\int \\frac{1}{(3 x+5)^{2}} d x$\n(vii) $\\int \\frac{2 x}{\\left(x^{2}+1\\right)} d x$\n(iv) $\\int 2 x e^{-x^{2}} d x$\n

Answer

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\neq- 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\neq-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\neq-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)