91Ó°ÊÓ

Evaluate the integrals that converge. $$\int_{0}^{1} \frac{d x}{\sqrt{x}(x+1)}$$

Evaluate the integral. $$\int \tan 4 x \sec ^{4} 4 x d x$$

Evaluate the integral. $$\int \tan ^{4} \theta \sec ^{4} \theta d \theta$$

Evaluate the integral. $$\int \frac{x^{3}+x^{2}+x+2}{\left(x^{2}+1\right)\left(x^{2}+2\right)} d x$$

(a) Derive the identity $$ \frac{\operatorname{sech}^{2} x}{1+\tanh ^{2} x}=\operatorname{sech} 2 x $$ (b) Use the result in part (a) to evaluate \(\int \operatorname{sech} x \, d x\) (c) Derive the identity $$ \operatorname{sech} x=\frac{2 e^{x}}{e^{2 x}+1} $$ (d) Use the result in part (c) to evaluate \(\int \operatorname{sech} x \, d x\) (e) Explain why your answers to parts (b) and (d) are consistent.

Evaluate the integral. $$\int_{0}^{\sqrt{3} / 2} \sin ^{-1} x d x$$

Evaluate the integrals that converge. $$\int_{0}^{+\infty} \frac{d x}{\sqrt{x}(x+1)}$$

The integral $$\int \frac{x^{2}}{x^{2}+4} d x$$ can be evaluated either by a trigonometric substitutior or by algebraically rewriting the numerator of the integrand as \(\left(x^{2}+4\right)-4 .\) Do it both ways and show that the results are equivalent.

Evaluate the integral. $$\int_{2}^{4} \sec ^{-1} \sqrt{\theta} d \theta$$

In Example 8 we showed that taking \(n=14\) subdivisions ensures that the approximation of $$\ln 2=\int_{1}^{2} \frac{1}{x} d x$$ by Simpson's rule is accurate to five decimal places. Confirm this by comparing the approximation of \(\ln 2\) produced by Simpson's rule with \(n=14\) to the value produced directly by your calculating utility.