Week 5 Activity

Section 5.4 Week 5 Activity

Subsection 5.4.1 Trigonometric Integrals

Activity 5.4.1.

Calculate this integral.

\begin{equation*} \int \sin^3 x \cos^4 x dx \end{equation*}

Solution.

The power of sine is odd. Therefore, if I seperate just \(\sin x\text{,}\) I should be able to setup a cosine substitution.

\begin{equation*} \int \sin^3 x \cos^4 x dx = \int \sin^2x \cos^4 x \sin x dx . \end{equation*}

Then I change everything into cosine, except the one seperate sine function.

\begin{align*} \amp = \int (1-\cos^2 x) \cos^4 x \sin x dx \end{align*}

Then I use a cosine substitution.

\begin{align*} u \amp = \cos x \\ du \amp = - \sin x dx \end{align*}

Using this substitution gives a polynomial integral. I can split it up using linearity and evalute each piece. When I am done that, I reverse the substitution.

\begin{align*} \int \sin x (1-\cos^2 x) \cos^4 x dx \amp = - \int (1-u^2)u^4 du \\ \amp = \int u^6 du - \int u^4 du = \frac{u^7}{7} - \frac{u^5}{5} + c \\ \amp = \frac{\cos^7 x}{7} - \frac{\cos^5 x}{5} + c \end{align*}

Activity 5.4.2.

Calculate this integral.

\begin{equation*} \int \sin^4 x \cos^2 x dx \end{equation*}

Solution.

Both powers are even, so a substitution will not work directly. Instead, I use double angle identities to reduce the exponents. Expansion of the resulting binomial terms produces four different integrals.

\begin{align*} \int \sin^4 x \cos^2 x dx \amp = \int \left( \frac{1 - \cos 2x}{2} \right)^2 \left( \frac{1 + \cos 2x}{2} \right) dx \\ \amp = \frac{1}{8} \int (1 - 2 \cos (2x) + \cos^2 (2x))(1+\cos (2x)) dx \\ \amp = \frac{1}{8} \int 1 - \cos (2x) - \cos^2 (2x) + \cos^3 (2x) dx \\ \amp = \frac{1}{8} \left[ \int 1 dx - \int \cos (2x) dx - \int \cos^2 (2x) dx + \int \cos^3 (2x) dx \right] \end{align*}

The first two integrals are done directly with known antiderivatives. The third needs another double angle identity. In the fourth, since the power is odd, I can seperate one cosine function, change everything else to sine, and use a sine substitution. The specific substitution here is \(u = \sin (2x)\) with \(du = 2 \cos (2x) dx \text{,}\) or \(\frac{1}{2} du = \cos (2x) dx\text{.}\)

\begin{align*} \amp = \frac{1}{8} \left[ x - \frac{\sin (2x)}{2} - \int \left( \frac{ 1 + \cos (4x)}{2} \right) dx + \int \cos^2 (2x) \cos (2x) dx \right] \\ \amp = \frac{1}{8} \left[ x - \frac{\sin (2x)}{2} - \frac{1}{2} \int 1 dx - \frac{1}{2} \int \cos (4x) dx + \int (1-\sin^2 (2x)) \cos (2x) dx \right]\\ \amp = \frac{1}{8} \left[ x - \frac{\sin (2x)}{2} - \frac{x}{2} - \frac{\sin (4x)}{8} + \frac{1}{2} \int (1-u^2) du \right] \\ \amp = \frac{1}{8} \left[ x - \frac{\sin (2x)}{2} - \frac{x}{2} - \frac{\sin (4x)}{8} + \frac{1}{2} \int du - \frac{1}{2} \int u^2 du \right] \end{align*}

From here, I sipmlify and group terms as I can. I also finish the \(u\) integral with known antiderivatives from the power rule and then reverse the substitution.

\begin{align*} \amp = \frac{1}{8} \left[ x - \frac{\sin (2x)}{2} - \frac{x}{2} - \frac{\sin (4x)}{8} + \frac{u}{2} - \frac{u^3}{6} \right] + c\\ \amp = \frac{1}{8} \left[ x - \frac{\sin (2x)}{2} - \frac{x}{2} - \frac{\sin (4x)}{8} + \frac{\sin (2x)}{2} - \frac{\sin^3 (2x)}{6} \right] + c\\ \amp = \frac{1}{8} \left[ \frac{x}{2} - \frac{\sin (4x)}{8} - \frac{\sin^3 (2x)}{6} \right] + c \end{align*}

Activity 5.4.3.

Calculate this integral.

\begin{equation*} \int \tan^3 x \sec^4 x dx \end{equation*}

Solution.

There is an even power of secant, so a tangent substitution should work well after isolating \(\sec^2 x\) and turning the remaining secant terms into tangent terms. The resulting \(u\) integral is a reasonable polynomial integral. After completing it, I reverse the substitution.

\begin{align*} \int \tan^3 x \sec^4 x dx \amp = \int \tan^3 x (1+\tan^2 x) \sec^2 x dx \\ u \amp = \tan x \implies du = \sec^2 x dx \\ \int \tan^3 x \sec^4 x dx \amp = \int u^3 (1+u^2) du = \int u^3 du + \int u^5 du \\ \amp = \frac{u^4}{4} + \frac{u^6}{6} + c = \frac{\tan^4}{4} + \frac{\tan^6}{6} + c \end{align*}

Activity 5.4.4.

Calculate this integral.

\begin{equation*} \int \sec^6 x dx \end{equation*}

Solution.

There is an even power of secant, so a tangent substitution should work well after isolateing \(\sec^2 x\) and turning the remaining secant terms into tangent terms. After the tangent substitution, the resulting integral is a polynomial integral. After completing the polynomial integral, I reverse the substitution.

\begin{align*} \int \sec^6 x dx \amp = \int (1+\tan^2 x)^2 \sec^2 x dx \\ u \amp = \tan x \implies du = \sec^2 x dx \\ \int \sec^6 x dx \amp = \int (1+u^2)^2 du = \int (1 + 2u^2 + u^4) du\\ \amp = \int du + 2 \int u^2 du + \int u^4 du \\ \amp = u + \frac{2u^3}{3} + \frac{u^5}{5} + c = \tan x + \frac{2 \tan^3 x}{3} + \frac{\tan^5 x}{5} + c \end{align*}

Activity 5.4.5.

Calculate this integral.

\begin{equation*} \int_0^{\frac{\pi}{3}} \tan x \sec^6 x dx \end{equation*}

Solution.

There is an even power of secant, so I can isolate \(\sec^2 x\) and turn the rest into tangent terms. Then a tangent substitution works. Since this is a definite integral, I’ll also change the bounds of the substitution, so that I can finish the polynomial integral with the new bounds and don’t have to do an inverse substitution.

\begin{align*} \int_0^{\frac{\pi}{3}} \tan x \sec^6 x dx \amp = \int_0^{\frac{\pi}{3}} \tan x (1+\tan^2 x)^2 \sec^2 x dx \\ u \amp = \tan x \implies du = \sec^2 x dx \\ x \amp = 0 \implies u = 0 \\ x \amp = \frac{\pi}{3} \implies u = \tan \frac{\pi}{3} = \sqrt{3}\\ \amp = \int_0^{\sqrt{3}} u(1+u^2)^2 du = \int_0^{\sqrt{3}} u(1+2u^2 + u^4) du\\ \amp = \int_0^{\sqrt{3}} u du + \int_0^{\sqrt{3}} 2u^3 du + \int_0^{\sqrt{3}} u^5 du\\ \amp = \frac{u^2}{2} + \frac{u^4}{2} + \frac{u^6}{6} \Bigg|_0^{\sqrt{3}} \\ \amp = \frac{3}{2} + \frac {9}{2} + \frac{27}{6} = \frac{63}{6} = \frac{31}{2} \end{align*}

Subsection 5.4.2 Trigonometric Substitutions

Activity 5.4.6.

Calculate this integral using a trig substitution. (Hint: after the substitution, write everything in terms of sine, break up the denominator, and then write \(\frac{1}{\sin \theta}\) as \(\csc \theta\text{.}\) Then you can use antiderivatives from the tables for both of the resulting trig integrals.)

\begin{equation*} \int \frac{\sqrt{9-x^2}}{x} dx \end{equation*}

Solution.

This is a sine substitution. I calculate the various pieces of the substitution.

\begin{align*} x \amp = 3 \sin \theta \\ dx \amp = 3 \cos \theta d \theta \\ \sqrt{9-x^2} \amp = \sqrt{9 - 9 \sin^2 \theta} = 3 \sqrt{1-\sin^2 \theta}\\ \amp = 3 \sqrt{\cos^2 \theta} = 3 \cos \theta \end{align*}

Then I use the pieces to change all the parts of the original integral into \(\theta\text{.}\) I simplify the result first by writing everything in terms of sine and then by split up the fraction and writing everything in the numerator.

\begin{align*} \int \frac{\sqrt{9-x^2}}{x} dx \amp = \int \frac{3 \cos \theta}{3 \sin \theta} 3 \cos \theta d \theta \\ \amp = 3 \int \frac{\cos^2 \theta}{\sin \theta} d \theta = 3 \int \frac{1 - \sin^2 \theta}{\sin \theta} d \theta \\ \amp = 3 \int \frac{1}{\sin theta} - \frac{\sin^2 \theta}{\sin \theta} d \theta = 3 \int \csc \theta d \theta - 3 \int \sin \theta d \theta \end{align*}

This is a bit unusual (as I noted in the hint). If I do the correct rearrangements, I don’t actually need any other substitution techniques; I can just use the antiderivatives for \(\sin \theta\) and \(\csc \theta\) from the tables.

\begin{equation*} 3 \int \csc \theta d \theta - 3 \int \sin \theta d \theta = 3 \ln |\csc \theta - \cot \theta| + 3 \cos \theta + c \end{equation*}

To reverse the substitution, I need to write everything in a form that I can substitute backward. I change the cosecant and cotangent back to sine and cosine. I also multiply and divide by \(3\) in both fractions, since my substitution pieces all have \(3\) in them.

\begin{align*} \amp = 3 \ln \left| \frac{1}{\sin \theta} - \frac{\cos \theta}{\sin \theta} \right| + 3 \cos \theta + c \\ \amp = 3 \ln \left| \frac{3}{3 \sin \theta} - \frac{ 3 \cos \theta}{3 \sin \theta} \right| + 3 \cos \theta + c \end{align*}

Now everything looks like something from the substitution calculations. I reverse the substitution.

\begin{equation*} \int \frac{\sqrt{9-x^2}}{x} dx = 3 \ln \left| \frac{3 - \sqrt{9-x^2}}{x} \right| + \sqrt{9-x^2} + c \end{equation*}

Activity 5.4.7.

Calculate this integral using a trig substitution. (Hint: after the substitution, you can write the sine terms as cosecant terms and look for a known antiderivative from the tables.)

\begin{equation*} \int \frac{1}{x^2 \sqrt{3-3x^2}} dx \end{equation*}

Solution.

I need to do a little algebra before this is in the correct form. I can take the \(3\) out of the square root (and also out of the integral, since it is constant).

\begin{align*} \int \frac{1}{x^2 \sqrt{3-3x^2}} dx \amp = \frac{1}{\sqrt{3}} \int \frac{1}{x^2 \sqrt{1-x^2}} dx \end{align*}

Now this is a sine substitution. I calculate the various pieces of the substitution.

\begin{align*} x \amp = \sin \theta \\ dx \amp = \cos \theta d\theta \\ \sqrt{1-x^2} \amp = \cos \theta \end{align*}

Now I replace all the pieces of the original integral with the trig substitution pieces. Then I can cancel out the cosine term.

\begin{align*} \frac{1}{\sqrt{3}} \int \frac{1}{x^2\sqrt{1-x^2}} dx \amp = \frac{1}{\sqrt{3}} \int \frac{1}{\sin^2 \theta \cos \theta} \cos \theta d\theta\\ \amp = \frac{1}{\sqrt{3}} \int \frac{1}{\sin^2 \theta} d\theta = \frac{1}{\sqrt{3}} \int \csc^2 \theta d \theta \end{align*}

I get the integral of cosecant squared. This happens to have a basic antiderivatives in the tables. After using this antiderivative, I write the result as sines and cosines to reverse the substitution. I will need to multiply and divide by \(3\) to make sure the expressions match the substitution. Then I reverse the substitution.

\begin{align*} \int \frac{1}{x^2 \sqrt{3-3x^2}} dx \amp = \frac{1}{\sqrt{3}} (- \cot \theta) + c = \frac{-\cos \theta}{\sqrt{3} \sin \theta} + c \\ \amp = \frac{-\cos \theta}{\sqrt{3} \sin \theta} + c = \frac{-\sqrt{1-x^2}}{\sqrt{3} x} + c \end{align*}

Activity 5.4.8.

Calculate this integral using a trig substitution.

\begin{equation*} \int \frac{\sqrt{x^2-6}}{x} dx \end{equation*}

Solution.

This is a secant substitution. I calculate the various piece of the substitution.

\begin{align*} x \amp = \sqrt{6} \sec \theta \\ dx \amp = \sqrt{6} \sec \theta \tan \theta d \theta \\ \sqrt{x^2-6} \amp = \sqrt{6} \tan \theta \\ \frac{x}{\sqrt{6}} \amp = \sec \theta \implies \arcsec \left( \frac{x}{\sqrt{6}} \right) = \theta \end{align*}

Then I replace all the piece of the original integral with the piece from the substitution. I can cancel of a secant and one \(\sqrt{6}\) to simplify.

\begin{align*} \int \frac{\sqrt{x^2-6}}{x} dx \amp = \int \frac{\sqrt{6} \tan \theta}{\sqrt{6} \sec \theta} \sqrt{6} \sec \theta \tan \theta d \theta \\ \amp = \sqrt{6} \int \tan^2 \theta d \theta \end{align*}

I can use the square identity for tangent and secant to write this as two reaonsable integrals (Remember that \(\sec^2 \theta\) has a nice antiderivative in the table.)

\begin{align*} \amp = \sqrt{6} \int \tan^2 \theta d \theta = \sqrt{6} \int (\sec^2 \theta - 1) d \theta \\ \amp = \sqrt{6} \int \sec^2 \theta d \theta - \sqrt{6} \int d\theta \\ \amp = \sqrt{6} \tan \theta - \sqrt{6} \theta + c \end{align*}

Then I can reverse the substitution. I have already calculated a term that I can use to replace the tangent. For replaing \(\theta\text{,}\) I invert the substitution.

\begin{equation*} \int \frac{\sqrt{x^2-6}}{x} dx = \sqrt{x^2-6} - \sqrt{6} \arcsec \left( \frac{x}{\sqrt{6}} \right) + c \end{equation*}

Activity 5.4.9.

Calculate this integral using a trig substitution.

\begin{equation*} \int_0^5 x^2 \sqrt{25-x^2} dx \end{equation*}

Solution.

This is a sine substitution. I calculate the various pieces of the substitution. Since this is a definite integral, I also calculate the new bounds. Note that calculating the new bounds involves inverting the sine function. To do this, I just used a unit circle diagram.

\begin{align*} x \amp = 5 \sin \theta \\ dx \amp = 5 \cos \theta d \theta \\ \sqrt{25-x^2} \amp = 5 \cos \theta \\ x \amp = 0 \implies 5 \sin \theta = 0 \implies \theta = 0 \\ x \amp = 5 \implies 5 \sin \theta = 5 \implies \sin \theta =1 \implies \theta = \frac{\pi}{2} \end{align*}

Then I do the substitution, replacing all the piece from the original as well as changing the bounds. I simplify by pulling out the constants.

\begin{align*} \int_0^5 x^2 \sqrt{25-x^2} dx \amp = \int_0^{\frac{\pi}{2}} 25 \sin^2 \theta 5 \cos \theta 5 \cos \theta d \theta\\ \amp = 625 \int_0^{\frac{\pi}{2}} \sin^2 \theta \cos^2 \theta d \theta \end{align*}

After substitution, there are even powers of both sine and cosine. I use the half-angle identities to reduce the exponents. That produces two integrals. (notice that two terms cancel when I multiply the binomials, leaving only two terms). In the second integral, I have to use the half-angle identities once again to deal with the \(\cos^2 (2\theta)\text{.}\) After doing so, all the remaining integrals have straightforward antiderivatives.

\begin{align*} \amp = 625 \int_0^{\frac{\pi}{2}} \left( \frac{1 - \cos (2\theta)}{2} \right) \left( \frac{1 + \cos (2\theta)}{2} \right) d \theta \\ \amp = \frac{625}{4} \int_0^{\frac{\pi}{2}} 1 - \cos^2 (2 \theta) d \theta \\ \amp = \frac{625}{4} \int_0^{\frac{\pi}{2}} d \theta - \frac{625}{4} \int_0^{\frac{\pi}{2}} \left( \frac{1 + \cos (4 \theta)}{2} \right) d \theta\\ \amp = \frac{625\pi}{8} - \frac{625}{8} \int_0^{\frac{\pi}{2}} d \theta - \frac{625}{8} \int_0^{\frac{\pi}{2}} \cos (4 \theta) d \theta \end{align*}

Since I changed the bounds, I don’t have to substitute back. I can simply evaluate on the new bounds.

\begin{align*} \amp = \frac{625\pi}{8} - \frac{625}{8} \frac{\pi}{2} - \frac{625}{8} \frac{-\sin 4 \theta}{4} \Bigg|_0^{\frac{\pi}{2}} \\ \amp = \frac{625\pi}{8} - \frac{625\pi}{16} - 0 = \frac{625\pi}{16} \end{align*}

Activity 5.4.10.

Calculate this integral using a trig substitution.

\begin{equation*} \int_0^{10} \frac{1}{\sqrt{100 + x^2}} dx \end{equation*}

Solution.

This is a tangent substitution. I calculate the various pieces of the substitution, including the bounds.

\begin{align*} x \amp = 10 \tan \theta \\ dx \amp = 10 \sec^2 \theta d \theta \\ \sqrt{100+x^2} \amp = \sec \theta \\ x \amp = 0 \implies 10 \tan \theta = 0 \implies \tan \theta = 0 \implies \theta = 0 \\ x \amp = 10 \implies 10 \tan \theta = 10 \implies \tan \theta = 1 \implies \theta = \frac{\pi}{4} \end{align*}

Then I perform the substitution, change the bounds, and simplify. One of the secant terms cancels. After that, I can use the tables to get the antiderivative of sectant.

\begin{align*} \int_0^{10} \frac{1}{\sqrt{100 + x^2}} dx \amp = \int_0^{\frac{\pi}{4}} \frac{1}{\sec \theta} \sec^2 \theta d \theta\\ \amp = \int_0^{\frac{\pi}{4}} \sec \theta d \theta \\ \amp = \ln | \tan \theta + \sec \theta| \Big|_0^{\frac{\pi}{4}} \end{align*}

Since I changed the bounds in the substitution, I don’t need a reverse substitution. I can just evaluate on these bounds.

\begin{align*} \amp = \ln \left| \tan \frac{\pi}{4} + \sec \frac{\pi}{4} \right| - \ln \left| \tan 0 + \sec 0 \right|\\ \amp = \ln | 1 + \sqrt{2} | - \ln |0 + 1| = \ln |1 + \sqrt{2}| \end{align*}

Subsection 5.4.3 Improper Integrals

Activity 5.4.11.

Calculate this improper integral and determine if it converges.

\begin{equation*} \int_1^\infty \frac{1}{x^3} dx \end{equation*}

Solution.

I find the antiderivative using the inverse power rule and then take the limit as the upper bound approaches \(\infty\text{.}\)

\begin{align*} \int_1^\infty \frac{1}{x^3} dx \amp = \lim_{a \rightarrow \infty} \int_1^a x^{-3} dx \\ \amp = \lim_{a \rightarrow \infty} \frac{x^{-2}}{-2} \Bigg|_1^a = \lim_{a \rightarrow \infty} \frac{1}{2} \left( \frac{-1}{a^2} + \frac{1}{1} \right) = \frac{1}{2} \end{align*}

This improper integral converges.

Activity 5.4.12.

Calculate this improper integral and determine if it converges. (You can do a substitution first and then the limit for the improper integral.)

\begin{equation*} \int_1^\infty \frac{\ln x}{x} dx \end{equation*}

Solution.

I use a substitution to solve the integral, including changing the bounds. I can deal with an infinite bound by determining how the new variable changes as the original variables goes to \(\infty\text{.}\)

\begin{align*} u \amp = \ln x \\ du \amp = \frac{1}{x} dx \\ x \amp = 1 \implies u = \ln 1 = 0 \\ x \amp \rightarrow \infty \implies u \rightarrow \infty \end{align*}

The last step in the substitution comes from the properties of the logarithm. If \(x \rightarrow \infty\text{,}\) then since the logarithm grows without bound, \(u\) will also diverge to infinity. Then I perform the substitution.

\begin{equation*} \int_1^\infty \frac{\ln x}{x} dx = \int_0^\infty u du \end{equation*}

Now I can use a limit to evalute this improper integral.

\begin{equation*} = \lim_{a \rightarrow \infty} \frac{u^2}{2} \Bigg|_0^a = \lim_{a \rightarrow \infty} \frac{a^2}{2} = \infty \end{equation*}

This improper integral diverges.

Activity 5.4.13.

Calculate this improper integral and determine if it converges.

\begin{equation*} \int_1^\infty x e^{-x} dx \end{equation*}

Solution.

I set up the improper integral limit.

\begin{equation*} \int_1^\infty x e^{-x} dx = \lim_{a \rightarrow \infty} \int_1^a x e^{-x} dx \end{equation*}

Then I use integration by parts.

\begin{align*} g(x) \amp = x \implies \frac{dg}{dx} = 1 \\ \frac{df}{dx} \amp = e^{-x} \implies f(x) = -e^{-x} \end{align*}

I need to bring the limit all the way through the process of integration by parts. The new integral after integration by parts is a reasonable exponential integral. I evaluate the first term and the antiderivative of the second integral on the bounds (using the termporary bound \(a\)).

\begin{align*} \amp = \lim_{a \rightarrow \infty} \left( -xe^{-x} \Big|_1^a + \int_1^a e^{-x} dx \right) = \lim_{a \rightarrow \infty} -ae^{-a} + e^{-1} + -e^{-x} \Big|_1^a\\ \amp = \lim_{a \rightarrow \infty} \frac{-a}{e^a} + \frac{1}{e} - \frac{1}{e^a} + \frac{1}{e} \end{align*}

Then I just need to evalute the limit. In the first term, the denominator with \(e^a\) clearly dominates over the numerator with just \(a\text{,}\) so the limit is zero.

\begin{equation*} \lim_{a \rightarrow \infty} \frac{-a}{e^a} + \frac{1}{e} - \frac{1}{e^a} + \frac{1}{e} = 0 + \frac{1}{e} + 0 + \frac{1}{e} = \frac{2}{e} \end{equation*}

This improper integral converges.

Activity 5.4.14.

Calculate this improper integral and determine if it converges.

\begin{equation*} \int_0^1 \ln x dx \end{equation*}

Solution.

The function is undefined at \(x=0\text{,}\) so I take a limit approaching this bound.

\begin{equation*} \int_0^1 \ln x dx = \lim_{a \rightarrow 0^+} \int_a^1 \ln x dx \end{equation*}

Then I just calculate the integral, using the antiderivative for the logarithm from the tables. After the antiderivative, I apply the limit.

\begin{equation*} \lim_{a \rightarrow 0^+} x \ln x - x \Bigg|_a^1 = \lim_{a \rightarrow 0^+} 1\ln 1 - 1 - a \ln a + a = -\infty \end{equation*}

This improper integral diverges, since as \(a\) gets close to zero, the logarithm \(\ln a\) diverges down to negative infinity. I makes sense that the area represented by the integral diverges to negative infinity, since the asymptote for the logarithm is in the negative direction. There is an infinite area below the \(x\) axis on this range, not above it.

Activity 5.4.15.

Calculate this improper integral and determine if it converges.

\begin{equation*} \int_0^1 \frac{1}{(x-1)^2} dx \end{equation*}

Solution.

The function is undefined at \(1\text{,}\) so we take a limit approaching this bound. Then I take the antiderivative with the inverse power rule (and, if I was being very careful and explicit, a substitution \(u = x-1\) to make this power rule work).

\begin{align*} \int_0^1 \frac{1}{(x-1)^2} dx \amp = \lim_{a \rightarrow 1^-} \int_0^a \frac{1}{(x-1)^2} dx \\ \amp = \lim_{a \rightarrow 1^-} \frac{-1}{x-1} \Bigg|_0^a = \lim_{a \rightarrow 1^-} \frac{-1}{a-1} + 1 = \infty \end{align*}

This improper integral diverges. (Note that I know the limit value will approach specificaly positive infinity. Since the limit approaches \(1\) from the negative, the denominator \(a-1\) will be negative. The numerator is also negative, so the quotient will be positive. Since the denominator is very close to zero, since the numerator is fixed, and since the quotient is always positive, the limit diverges to infinity.

Activity 5.4.16.

Calculate this improper integral and determine if it converges.

\begin{equation*} \int_0^{\frac{\pi}{2}} \tan x dx \end{equation*}

Solution.

The function is undefined at \(\frac{\pi}{2}\text{,}\) so I take a limit approaching this bound. I use the antiderivative of tangent from the tables.

\begin{align*} \int_0^{\frac{\pi}{2}} \tan x dx \amp = \lim_{a \rightarrow \frac{\pi}{2}^-} \int_0^a \tan x dx \\ \amp = \lim_{a \rightarrow \frac{\pi}{2}^-} -\ln \cos x \Bigg|_0^a = \lim_{a \rightarrow \frac{\pi}{2}^-} -\ln \cos a + \ln \cos 0 = \infty \end{align*}

This improper integral diverges. I know the limit is divergent since, as \(a \rightarrow \frac{\pi}{2}\text{,}\) cosine approaches 0 from the positive side, and the logarithm of something approaching 0 from the positive side is approaching \(-\infty\text{.}\)

Activity 5.4.17.

Calculate this improper integral using a trig substitution first and then the improper integral limit after the trig substitution.

\begin{equation*} \int_{2\sqrt{2}}^{4} \frac{1}{\sqrt{(16-x^2)^3}} dx \end{equation*}

Solution.

Even though this is an improper integral (it is undefined at \(x=4\)), I can do the trig substitution first and then do the limit for the improper integral. This is a sine substitution. I calculate the various pieces of the substitution, including the bounds.

\begin{align*} x \amp = 4 \sin \theta \\ dx \amp = 4 \cos \theta d \theta \\ \sqrt{16-x^2} \amp = 4 \cos \theta \\ x \amp = 4 \implies 4 \sin \theta = 4 \implies \sin \theta = 1 \implies \theta = \frac{\pi}{2}\\ x \amp = 2\sqrt{2} \implies 4 \sin \theta = 2\sqrt{2} \implies \sin \theta = \frac{1}{\sqrt{2}} \implies \theta = \frac{\pi}{4} \end{align*}

Then I perform the substitution and simplify. This nicely simplifies down to \(\sec^2 \theta\text{,}\) which has a known antiderivative from the tables.

\begin{align*} \int_{2\sqrt{2}}^{4} \frac{1}{\sqrt{(16-x^2)^3}} dx \amp = \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \frac{1}{64 \cos^3 \theta} 4 \cos \theta d \theta\\ \amp = \frac{1}{16}\int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \frac{1}{\cos^2 \theta} d \theta\\ \amp = \frac{1}{16}\int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \sec^2 \theta d \theta \\ \amp = \frac{1}{16} \tan \theta \Big|_{\frac{\pi}{4}}^{\frac{\pi}{2}} \end{align*}

Now I’ll to the improper integral. Tangent is undefined at \(\frac{\pi}{2}\text{,}\) so I can the limit approaching that endpoint.

\begin{equation*} \frac{1}{16} \tan \theta \Big|_{\frac{\pi}{4}}^{\frac{\pi}{2}} = \lim_{a \rightarrow \frac{\pi}{2}} \frac{1}{16} \tan \theta \Big|_{\frac{\pi}{4}}^{a} = \lim_{a \rightarrow \frac{\pi}{2}} \frac{1}{16} \left( \tan a - 1 \right) = \infty \end{equation*}

The limit diverges since I know that tangent has a vertical asymptote at \(\frac{\pi}{2}\text{.}\) The improper integral does not converge, and I conclude the area under the original function on the given range was infinite.

Subsection 5.4.4 Conceptual Review Questions

How are various trig identities useful in integration strategies?
Why do trig substitution remove square roots?
Why do limit allow us to consider improper integrals?

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