Question

1. (exam 4 review 5b) solve the initial value problem: $g(t)=\frac{-}{t^{2}}$; $g(-1)=g(1)=1$.
2. (exam 4 review 14) evaluate the integral: $\int\frac{t^{2}\sec^{2}t^{3}}{(\tan t^{3}-1)^{2}}dt$

Explanation:

Step1: Integrate \(g''(t)\) to find \(g'(t)\)

We know that \(g''(t)=\frac{- 2}{t^{2}}=-2t^{-2}\).
Using the power - rule for integration \(\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C\) (\(n
eq - 1\)), we have:
\(g'(t)=\int g''(t)dt=\int-2t^{-2}dt\)
\(g'(t)=-2\times\frac{t^{-2 + 1}}{-2+1}+C\)
\(g'(t)=\frac{2}{t}+C\)

Step2: Use the initial condition \(g'(-1) = 1\) to find \(C\)

Substitute \(t=-1\) and \(g'(-1) = 1\) into \(g'(t)=\frac{2}{t}+C\)
\(1=\frac{2}{-1}+C\)
\(1=-2 + C\), so \(C = 3\)
Then \(g'(t)=\frac{2}{t}+3\)

Step3: Integrate \(g'(t)\) to find \(g(t)\)

\(g(t)=\int g'(t)dt=\int(\frac{2}{t}+3)dt\)
Using the rules \(\int\frac{1}{x}dx=\ln|x|+C\) and \(\int kdx=kx + C\) (\(k\) is a constant), we get:
\(g(t)=2\ln|t|+3t+D\)

Step4: Use the initial condition \(g(1) = 1\) to find \(D\)

Substitute \(t = 1\) and \(g(1)=1\) into \(g(t)=2\ln|t|+3t+D\)
\(1=2\ln(1)+3\times1+D\)
Since \(\ln(1)=0\), we have \(1 = 0+3+D\)
\(D=-2\)

Answer:

\(g(t)=2\ln|t|+3t - 2\)