Problems

1. For each of the following integrals, graph the integrand on the interval of integration. Use the graph to make a visual estimate of the integral, preferably one of the form "It has to be at least this big, and it can't be any bigger than that." If you have an exact answer from Exercise 2, check that it fits with your estimate. Otherwise, use the Definite Integral tool to check your estimate. If your estimate is not consistent with the actual value, check your calculation for possible errors. (Note that both tools expect functions of \(x\).)

   \(\displaystyle\int_0^1 x^2 \sqrt{1+x^3}\;dx\)	\(\displaystyle\int_0^1\,t\,e^{-\frac{t^2}{2}} \,dt\)	\(\displaystyle\int_0^{\,\pi}\,\sin^2 \theta\, d \theta\)
   \(\displaystyle\int_0^1\,x \sqrt{x^2+2}\;dx\)	\(\displaystyle\int_0^{\,\pi}\,\cos^2 x\, \sin\,x\,dx\)	\(\displaystyle\int_0^1\,x \sqrt{9-x^2} \;dx\)
   \(\displaystyle\int_0^1\,\frac{x}{1+x^2}\,dx\)	\(\displaystyle\int_0^1\,\frac{x}{\sqrt{1+x^2}} \,dx\)	\(\displaystyle\int_{-1}^{\,1} \,\sqrt{1-\frac{x}{4}} \;dx\)
   \(\displaystyle\int_0^1 \,\frac{x}{\sqrt{9-x^2}}\;dx\)	\(\displaystyle\int_0^1\,\sqrt{9-x^2}\;dx\)	\(\displaystyle\int_0^1\,\frac{x^2}{\sqrt{9-x^2}}\;dx\)
2. We observed in Example 1 that a substitution of the form \(x=a\,\sin\,\theta\) would turn an expression of the form \(\sqrt{a^2-x^2}\) into a perfect square. (The value of \(a\) was \(5\) in our example.)
   1. Show that a substitution of the form \(x=a\,\tan\,\theta\) will turn an expression of the form \(\sqrt{a^2+x^2}\) into a perfect square.
   2. Use the idea in part (a) to show that

      \(\displaystyle\int\; \frac{dx}{\left(25+x^2\right)^{3/2}}\)

      can be transformed into

      \(\displaystyle\frac{1}{25} \int\; \cos\,\theta\,d \theta = \frac{1}{25} \sin\,\theta + C\).

   3. The antidifferentiation in part (b) isn't finished until we express the answer in terms of \(x\). Draw a right triangle with sides \(x\) and \(5\), and label the angle \(\theta\) in this triangle for which \(x=5\,\tan\,\theta\). What is the length of the hypotenuse of your triangle? What is \(\sin\,\theta\) in terms of \(x\)? Complete your calculation of the indefinite integral.
   4. Check your answer to part (c) by differentiation.
3. 1. Apply the substitution of the preceding exercise to \(\displaystyle\int\,\frac{dx}{\sqrt{25+x^2}}\). What transformed integral do you get? Are you any closer to solution of the original problem?
   2. Calculate the derivative of the function \(\ln|\sec\,\theta+\tan\,\theta\,|\). Now can you evaluate the transformed integral in part (a)? Do it.
   3. Complete the original problem.
4. In Example 1 we argued that \(\sqrt{\cos^2\,\theta}=\cos\,\theta\) (with no absolute value needed) whenever we make a trigonometric substitution of the form \(x=a\,\sin\,\theta\), because each value of \(\theta\) is a value of the arcsine function, and therefore lies between \(-\pi/2\) and \(\pi/2\). (We also implicitly assumed that \(a\) is the positive square root of \(a^2\) - an assumption we are always free to make.) In Exercise 5 we introduced the trigonometric substitution  \(x=a\,\tan\,\theta\) and blithely assumed that \(\sqrt{\sec^2\,\theta}=\sec\,\theta\) (again with no absolute value). Explain why \(\sec\,\theta\) is always nonnegative for the relevant values of \(\theta\).
5. Evaluate each of the following integrals.

   a.   \(\displaystyle\int\,\frac{1}{9+x^2} \,dx\)

   b.   \(\displaystyle\int_0^1 \frac{1}{9+x^2}\,dx\)

   c.   \(\displaystyle\int_0^1 \frac{1}{7+x^2} \,dx\)

6. Use integration by parts, with \(u=\tan^{-1}\,x\) and \(dv=2x\,dx\), to evaluate \(\displaystyle\int\,2x\,\tan^{-1}\,x\;dx\) two ways:
   1. with \(v=x^2\), and
   2. with \(v=x^2+1\).
   How much difference does it make to choose your antidifferential of \(dv\) carefully?
7. Evaluate each of the following integrals.

   a.   \(\displaystyle\int_0^1 x^2 \sqrt{4x^3+2}\,dx\)	b.   \(\displaystyle\int\,\left(t^2+\sin\,3t\right)\,dt\)	c.   \(\displaystyle\int\,\frac{7}{\sqrt{s+1}}\;ds\)
   d.   \(\displaystyle\int\,\frac{1}{x^2+4} \,dx\)	e.   \(\displaystyle\int_3^{\,7}\, \frac{4}{2y+1}\,dy\)	f.   \(\displaystyle\int_{-\pi/4}^{\,\pi/4} \,\cos\,2x\,dx\)
   g.   \(\displaystyle\int_0^2\, e^{-0.3t} \,dt\)	h.  \(\displaystyle\int\, 2x\left(1+x^2\right)^{3/4}dx\)	i.   \(\displaystyle\int\,\sqrt{1-4w^2} \,dw\)
   j.   \(\displaystyle\int_0^{\,\pi/4} \,\sin^2\,6 \theta\,d \theta\)	k.   \(\displaystyle\int_{-\pi}^{\,\pi}\,\cos^2\,3t\,dt\)	l.   \(\displaystyle\int\,t\,e^{5.3t} \,dt\)
   m.   \(\displaystyle\int_0^{\,1}\,t\, e^{5.3t}\,dt\)	n.   \(\displaystyle\int\,(t+3)\,e^{5.3t} \,dt\)	o.   \(\displaystyle\int\,t\,\cos\,2t\,dt\)
   p.   \(\displaystyle\int\, t^2\,\sin\,2t\,dt\)	q.   \(\displaystyle\int_0^{\pi/2}\,t^2\, \sin\,2t\,dt\)	r.   \(\displaystyle\int_0^{\,1}\,\frac{x}{\left(1+x^2\right)^2}\,dx\)

8. 1. Guess which of the following integrals will be larger, and explain your reasoning:

      \(\displaystyle\int_0^{\,4}\,x \sqrt{16-x^2}\;dx\)    or    \(\displaystyle\int_0^{\,4}\, \sqrt{16-x^2}\;dx\)

   2. Compute which of the two integrals is actually larger.

Problem adapted from Calculus Problems for a New Century, edited by Robert Fraga, MAA Notes Number 28, 1993.

9. In Example 2 we suggested evaluating \(\displaystyle\int_0^{\,\pi}\,\cos^2\,\theta\,d \theta\) by using a double-angle formula to transform the integrand. Suppose instead that we make a substitution: \(u=\sin\,\theta\). Then \(\cos\,\theta=\sqrt{1-u^2}\), \(du=\cos\,\theta\,d \theta\), and \(u=0\) at both \(\theta=0\) and \(\theta=\pi\). That would seem to transform the integral into \(\displaystyle\int_0^{\,0}\,\sqrt{1-u^2} \;du\), which is clearly \(0\). But the original function being integrated has only non-negative values, so the correct value of the integral must be positive. What went wrong? What is the correct value of the integral?

Problem adapted from Calculus Problems for a New Century, edited by Robert Fraga, MAA Notes Number 28, 1993.