91Ó°ÊÓ

Find the arc length of the curve \(y=\ln x\) from \(x=1\) to \(x=2.\)

(a) Derive the identity $$ \frac{\sec ^{2} x}{\tan x}=\frac{1}{\sin x \cos x} $$ (b) Use the identity \(\sin 2 x=2 \sin x \cos x\) along with the result in part (a) to evaluate \(\int \csc x \, d x\) (c) Use the identity \(\cos x=\sin [(\pi / 2)-x]\) along with your answer to part (a) to evaluate \(\int \sec x \, d x\)

(a).Make the indicated \(u\) -substitution, and then use the End paper Integral Table to evaluate the integral. (b).If you have a CAS, use it to evaluate the integral, and then confirm that the result is equivalent to the one that you found in part (a). $$\int \frac{\sin ^{2}(\ln x)}{x} d x, u=\ln x$$

Evaluate the integral. $$\int \frac{x^{3}-2 x^{2}+2 x-2}{x^{2}+1} d x$$

Evaluate the integral. $$\int \sec ^{5} x \tan ^{3} x d x$$

Determine whether the statement is true or false. Explain your answer. If \(f\) is continuous on \([a,+\infty)\) and \(\lim _{x \rightarrow+\infty} f(x)=1,\) then \(\int_{a}^{+\infty} f(x) d x\) converges.

Find the arc length of the curve \(y=x^{2}\) from \(x=0\) to \(x=1\)

Evaluate the integral. $$\int \tan ^{5} \theta \sec \theta d \theta$$

In each part, determine whether a trapezoidal approximation would be an underestimate or an overestimate for the definite integral. $$\text { (a) } \int_{0}^{1} \cos \left(x^{2}\right) d x$$ $$\text { (b) } \int_{3 / 2}^{2} \cos \left(x^{2}\right) d x$$

Evaluate the integral. $$\int_{1}^{2} x \sec ^{-1} x d x$$