Question
For $1< B < \infty,$ compute $I(B)=\int_{1}^{B} \frac{d t}{t \sqrt{t^{2}-1}}$ and evaluate $\lim _{B \rightarrow \infty} I(B), \quad$ the area under the graph of $\frac{1}{t \sqrt{t^{2}-1}}$ over $[1, \infty)$
Step 1
This integral can be solved by using the substitution $t=\cosh(u)$, which gives $dt=\sinh(u) du$ and $t^2-1=\sinh^2(u)$. The integral then becomes $\int \frac{du}{\cosh(u)}$, which is equal to $\text{arccosh}(t)$. Show more…
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