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 Trig Integration Trig Substitution Integration by Parts L'Hopitals Rule Improper Integrals

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17calculus > exam list > calc2 exam C2

This is the second of four exams for semester C of calculus 2.

### Practice Exam Tips

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IMPORTANT -
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Calculus is not something that can be learned by reading. You have to work problems on your own and struggle through the material to really know calculus, do well on your exam and be able to use it in the future.

Exam Details

Time

2.5 hours

Questions

7+1EC

Total Points

85+10EC

Tools

Calculator

no

Formula Sheet(s)

1 page, 8.5x11 or A4

Other Tools

ruler for drawing graphs

Instructions:
- For each problem, correct answers are worth 1 point. The remaining points are earned by showing calculations and giving reasoning that justify your conclusions.
- Correct notation counts (i.e. points will be taken off for incorrect notation).

[10 points] Evaluate either integral [a] or [b] of this question. No extra credit for working both. Turn in work for only one.
[a] $$\int{ \sin^4\theta \cos^5\theta ~d\theta }$$
[b] $$\int{ \tan^2\theta \sec^4\theta ~d\theta }$$

Problem Statement

[10 points] Evaluate either integral [a] or [b] of this question. No extra credit for working both. Turn in work for only one.
[a] $$\int{ \sin^4\theta \cos^5\theta ~d\theta }$$
[b] $$\int{ \tan^2\theta \sec^4\theta ~d\theta }$$

[a] $$\displaystyle{ \int{ \sin^4\theta \cos^5\theta ~d\theta } = \frac{\sin^5\theta}{5} - \frac{2\sin^7\theta}{7} + \frac{\sin^9\theta}{9} + C }$$
[b] $$\displaystyle{ \int{ \tan^2\theta \sec^4\theta ~d\theta } = \frac{\tan^3\theta}{3} + \frac{\tan^5\theta}{5} + C }$$

Problem Statement

[10 points] Evaluate either integral [a] or [b] of this question. No extra credit for working both. Turn in work for only one.
[a] $$\int{ \sin^4\theta \cos^5\theta ~d\theta }$$
[b] $$\int{ \tan^2\theta \sec^4\theta ~d\theta }$$

Solution

 [a] $$\int{ \sin^4\theta \cos^5\theta ~d\theta }$$ $$\int{ \sin^4\theta (\cos^2\theta)^2 \cos\theta ~d\theta }$$ Use $$\cos^2\theta = 1-\sin^2\theta$$ $$\int{ \sin^4\theta (1-\sin^2\theta)^2 \cos\theta ~d\theta }$$ Use integration by substitution. $$u=\sin\theta \to du=\cos\theta~d\theta$$ $$\int{ u^4(1-u^2)^2 ~du }$$ $$\int{ u^4(1-2u^2+u^4) ~du }$$ $$\int{ u^4-2u^6+u^8 ~du }$$ $$\displaystyle{ \frac{u^5}{5} - \frac{2u^7}{7} + \frac{u^9}{9} + C }$$ $$\displaystyle{ \frac{\sin^5\theta}{5} - \frac{2\sin^7\theta}{7} + \frac{\sin^9\theta}{9} + C }$$
 Final Answer - Integral [a] $$\displaystyle{ \int{ \sin^4\theta \cos^5\theta ~d\theta } = \frac{\sin^5\theta}{5} - \frac{2\sin^7\theta}{7} + \frac{\sin^9\theta}{9} + C }$$
 [b] $$\int{ \tan^2\theta \sec^4\theta ~d\theta }$$ $$\int{ \tan^2\theta \sec^2\theta \sec^2\theta ~d\theta }$$ Use $$1+\tan^2\theta = \sec^2\theta$$ $$\int{ \tan^2\theta (1+\tan^2\theta) \sec^2\theta ~d\theta }$$ Use integration by substitution. $$u=\tan\theta \to du=\sec^2\theta~d\theta$$ $$\int{ u^2(1+u^2) ~du }$$ $$\int{ u^2 + u^4 ~du }$$ $$\displaystyle{ \frac{u^3}{3} + \frac{u^5}{5} + C }$$ $$\displaystyle{ \frac{\tan^3\theta}{3} + \frac{\tan^5\theta}{5} + C }$$
 Final Answer - Integral [b] $$\displaystyle{ \int{ \tan^2\theta \sec^4\theta ~d\theta } = \frac{\tan^3\theta}{3} + \frac{\tan^5\theta}{5} + C }$$

[a] $$\displaystyle{ \int{ \sin^4\theta \cos^5\theta ~d\theta } = \frac{\sin^5\theta}{5} - \frac{2\sin^7\theta}{7} + \frac{\sin^9\theta}{9} + C }$$
[b] $$\displaystyle{ \int{ \tan^2\theta \sec^4\theta ~d\theta } = \frac{\tan^3\theta}{3} + \frac{\tan^5\theta}{5} + C }$$

[10 points] Evaluate $$\displaystyle{ \int{ \frac{dx}{x\sqrt{16-9x^2}} } }$$ using trig substitution.

Problem Statement

[10 points] Evaluate $$\displaystyle{ \int{ \frac{dx}{x\sqrt{16-9x^2}} } }$$ using trig substitution.

$$\displaystyle{ \frac{1}{4}\ln|x| - \frac{1}{4}\ln|4+\sqrt{16-9x^2}| + C }$$

Problem Statement

[10 points] Evaluate $$\displaystyle{ \int{ \frac{dx}{x\sqrt{16-9x^2}} } }$$ using trig substitution.

Solution

 $$3x = 4\sin\theta \to dx=(4/3)\cos\theta~d\theta$$ Based on the equation $$3x/4 = \sin\theta$$, draw a triangle as shown with $$a=3x$$, $$c=4$$ and $$b=\sqrt{16-9x^2}$$. Substitute the trig terms into the integral. $$\displaystyle{ \int{ \frac{dx}{x\sqrt{16-9x^2}} } }$$ $$\displaystyle{ \int{ \frac{(4/3)\cos\theta ~d\theta}{(4/3)\sin\theta\sqrt{16-16\sin^2\theta}} } }$$ $$\displaystyle{ \int{ \frac{\cos\theta ~d\theta}{4\sin\theta\cos\theta} } }$$ $$\displaystyle{ \frac{1}{4} \int{ \frac{1}{\sin\theta} ~d\theta } }$$ $$\displaystyle{ \frac{1}{4}\int{ \csc\theta ~d\theta } }$$ $$\displaystyle{ \frac{-1}{4}\ln |\csc\theta+\cot\theta | + C }$$ From the triangle that we built, we can write $$\csc\theta = 4/(3x)$$ and $$\cot\theta = \sqrt{16-9x^2}/(3x)$$ and substitute these into the answer involving $$\theta$$. $$\displaystyle{ \frac{-1}{4}\ln \left| \frac{4}{3x} + \frac{\sqrt{16-9x^2}}{3x} \right| + C_1}$$ Depending on what your instructor expects as far as simplification, the following answers are also correct. $$\displaystyle{ \frac{-1}{4}\ln \left| \frac{4+\sqrt{16-9x^2}}{3x} \right| + C_1}$$ $$\displaystyle{ \frac{1}{4}\ln \left| \frac{3x}{4+\sqrt{16-9x^2}} \right| + C_1}$$ $$\displaystyle{ \frac{1}{4}\ln|3x| - \frac{1}{4}\ln|4+\sqrt{16-9x^2}| + C_1 }$$ $$\displaystyle{ \frac{1}{4}\ln|x| - \frac{1}{4}\ln|4+\sqrt{16-9x^2}| + C_2 }$$

In order to get the last answer, we write $$(1/4)\ln|3x|$$ as $$(1/4)[ \ln(3) + \ln|x|]$$. Then we absorb the constant $$(1/4)\ln(3)$$ into the constant $$C_1$$ to get $$C_2 = C_1 + (1/4)\ln(3)$$.
A similar procedure could be done on each of the previous answers. We believe that last answer is the simplest.
As far the absolute value signs go, they are required on the $$\ln|3x|$$ or $$\ln|x|$$ term since we are not guaranteed by the integral in the question that $$x \gt 0$$. However, absolute value signs are not required on the term $$\ln|4+\sqrt{16-9x^2}|$$ since the $$4+\sqrt{16-9x^2}$$ is always positive and, in fact, it is greater than or equal to 4. That said, leaving them absolute value signs on is correct.

$$\displaystyle{ \frac{1}{4}\ln|x| - \frac{1}{4}\ln|4+\sqrt{16-9x^2}| + C }$$

[10 points] Evaluate $$\displaystyle{ \lim_{x \to \infty}{ \frac{x^2}{(\ln x)^3} } }$$

Problem Statement

[10 points] Evaluate $$\displaystyle{ \lim_{x \to \infty}{ \frac{x^2}{(\ln x)^3} } }$$

$$\displaystyle{ \lim_{x \to \infty}{ \frac{x^2}{(\ln x)^3} } = \infty }$$

Problem Statement

[10 points] Evaluate $$\displaystyle{ \lim_{x \to \infty}{ \frac{x^2}{(\ln x)^3} } }$$

Solution

 Start by trying direct substitution. $$\displaystyle{ \lim_{x \to \infty}{ \frac{x^2}{(\ln x)^3} } = \frac{\infty}{\infty} }$$ This is indeterminate, so apply L'Hopitals Rule. $$\displaystyle{ \lim_{x \to \infty}{ \frac{2x}{3(\ln x)^2(1/x)} } }$$ $$\displaystyle{ \lim_{x \to \infty}{ \frac{2x^2}{3(\ln x)^2} } = \frac{\infty}{\infty} }$$ Indeterminate again, so repeat L'Hopitals Rule. $$\displaystyle{ \lim_{x \to \infty}{ \frac{4x}{6(\ln x)(1/x)} } }$$ $$\displaystyle{ \lim_{x \to \infty}{ \frac{2x^2}{3(\ln x)} } = \frac{\infty}{\infty} }$$ Indeterminate again, so repeat L'Hopitals Rule a third time. $$\displaystyle{ \lim_{x \to \infty}{ \frac{4x}{3(1/x)} } }$$ $$\displaystyle{ \lim_{x \to \infty}{ \frac{4x^2}{3} } = \infty }$$

Notice that we can use L'Hopitals Rule multiple times as long as we use it only an indeterminate form. So each time, we need to check.

$$\displaystyle{ \lim_{x \to \infty}{ \frac{x^2}{(\ln x)^3} } = \infty }$$

[15 points] [a] Use integration by parts to show that $$\int{\ln x~dx} = x\ln x -x +C$$.
[b] Integrate $$\int{x^2\ln x~dx}$$.

Problem Statement

[15 points] [a] Use integration by parts to show that $$\int{\ln x~dx} = x\ln x -x +C$$.
[b] Integrate $$\int{x^2\ln x~dx}$$.

[b] $$\displaystyle{ \frac{x^3}{3} \left[ 3\ln x - 1 \right] + C }$$

Problem Statement

[15 points] [a] Use integration by parts to show that $$\int{\ln x~dx} = x\ln x -x +C$$.
[b] Integrate $$\int{x^2\ln x~dx}$$.

Solution

 [a] $$\int{\ln x~dx}$$ As directed, use integration by parts. $$u = \ln x \to du = (1/x)dx$$ $$dv = dx \to v=x$$ $$x\ln x - \int{ x (1/x) ~dx }$$ $$x\ln x - \int{ ~dx }$$ $$x\ln x - x + C$$
 [b] $$\int{x^2\ln x~dx}$$ Use integration by parts. $$u = \ln x \to du = (1/x)dx$$ $$dv = x^2~dx \to v = x^3/3$$ $$\displaystyle{ \frac{x^3}{3}\ln x - \int{ \frac{x^3}{3}\frac{1}{x}~dx } }$$ $$\displaystyle{ \frac{x^3}{3}\ln x - \frac{1}{3} \int{ x^2 ~dx } }$$ $$\displaystyle{ \frac{x^3}{3}\ln x - \frac{1}{3}\frac{x^3}{3} + C }$$ $$\displaystyle{ \frac{x^3}{3}\ln x - \frac{x^3}{9} + C }$$ $$\displaystyle{ \frac{x^3}{3} \left[ 3\ln x - 1 \right] + C }$$

[b] $$\displaystyle{ \frac{x^3}{3} \left[ 3\ln x - 1 \right] + C }$$

[15 points] Evaluate $$\displaystyle{ \lim_{x\to 0^+}{ \left[ \frac{1}{\ln(x+1)} - \frac{1}{x} \right] } }$$

Problem Statement

[15 points] Evaluate $$\displaystyle{ \lim_{x\to 0^+}{ \left[ \frac{1}{\ln(x+1)} - \frac{1}{x} \right] } }$$

$$\displaystyle{ \lim_{x\to 0^+}{ \left[ \frac{1}{\ln(x+1)} - \frac{1}{x} \right] } = \frac{1}{2} }$$

Problem Statement

[15 points] Evaluate $$\displaystyle{ \lim_{x\to 0^+}{ \left[ \frac{1}{\ln(x+1)} - \frac{1}{x} \right] } }$$

Solution

Direct substitution yields $$\infty - \infty$$ which is indeterminate. We would like to use L'Hopitals Rule but we can only do L'Hopitals Rule on fractions. So we need to do some algebra to get the equation in the right form. So, we get a common denominator and combine the terms into one fraction.

 $$\displaystyle{ \lim_{x\to 0^+}{ \left[ \frac{x-\ln(x+1)}{x\ln(x+1)} \right] } }$$

Checking this, we get $$0/0$$ which is still indeterminate. So apply L'Hopitals Rule.

 $$\displaystyle{ \lim_{x\to 0^+}{ \left[ \frac{1-1/(x+1)}{x(1/(x+1))+\ln(x+1)} \right] } }$$ $$\displaystyle{ \lim_{x\to 0^+}{ \left[ \frac{x}{x+(x+1)\ln(x+1)} \right] } }$$

Direct substitution yields $$0/0$$, still indeterminate. Since it looks like we may be making progress, let's apply L'Hopitals Rule again.

 $$\displaystyle{ \lim_{x\to 0^+}{ \left[ \frac{1}{1+(x+1)(1/(x+1))+\ln(x+1)} \right] } }$$ $$\displaystyle{ \lim_{x\to 0^+}{ \left[ \frac{1}{2+\ln(x+1)} \right] } }$$

This time, direct substitution gives us $$1/2$$.

$$\displaystyle{ \lim_{x\to 0^+}{ \left[ \frac{1}{\ln(x+1)} - \frac{1}{x} \right] } = \frac{1}{2} }$$

[15 points] Evaluate $$\displaystyle{ \int_0^1{x\ln x~dx} }$$.

Problem Statement

[15 points] Evaluate $$\displaystyle{ \int_0^1{x\ln x~dx} }$$.

$$-1/4$$

Problem Statement

[15 points] Evaluate $$\displaystyle{ \int_0^1{x\ln x~dx} }$$.

Solution

This integral can't be evaluated directly because there is a discontinuity in the lower limit, i.e. at $$x=0$$, $$\ln(x)$$ does not exist. Integration can be done only on continuous closed intervals. This is an improper integral. To make it proper, write it as $$\displaystyle{ \lim_{a\to 0^+}{\int_a^1{x\ln x~dx}} }$$.
We must write this as $$a\to 0^+$$ since we are working on the right side of zero.
We will now drop the limits of integration and work with the indefinite integral in order to not have to carry along so much notation. Then we will go back and apply the limits.

 $$\int{x\ln x~dx}$$ Use integration by parts. $$u = \ln x \to du = (1/x)dx$$ $$dv = x~dx \to v = x^2/2$$ $$\displaystyle{ \frac{x^2}{2}\ln x - \int{ \frac{x^2}{2}\frac{1}{x}~dx } }$$ $$\displaystyle{ \frac{x^2}{2}\ln x - \frac{1}{2} \int{ x ~dx } }$$ $$\displaystyle{ \frac{x^2}{2}\ln x - \frac{1}{2} \frac{x^2}{2} }$$ $$\displaystyle{ \frac{x^2}{2}\ln x - \frac{x^2}{4} }$$

Now apply the limits.

 $$\displaystyle{ \lim_{a\to 0^+}{\int_a^1{x\ln x~dx}} }$$ $$\displaystyle{ \lim_{a\to 0^+}{ \left[ \frac{x^2}{2}\ln x - \frac{x^2}{4} \right]_a^1 } }$$ $$\displaystyle{ \lim_{a\to 0^+}{ \left[ \frac{1}{2}\ln 1 - \frac{1}{4} \right] } - \lim_{a\to 0^+}{ \left[ \frac{a^2}{2}\ln a - \frac{a^2}{4} \right] } }$$

The first limit is pretty easy. Since $$\ln 1 = 0$$ we are left with $$-1/4$$.
In the second limit, the second term is zero. So we are left with the first term.

 $$\displaystyle{ \lim_{a\to0^+}{ a^2\ln a } = 0\cdot\infty }$$

This is indeterminate. So we need to rewrite the equation into a fraction, so that we can use L'Hopitals Rule.

 $$\displaystyle{ \lim_{a\to0^+}{ \frac{\ln a}{a^{-2}} } }$$ Apply L'Hopitals Rule. $$\displaystyle{ \lim_{a\to0^+}{ \frac{1/a}{-2a^{-3}} } }$$ $$\displaystyle{ \lim_{a\to0^+}{ \frac{-a^2}{2} = 0 } }$$

So we end up with $$-1/4$$ for the limit and for the improper integral.

$$-1/4$$

[10 points] For the following improper integral, set up but do not evaluate, limits and integrals such that the integrals can be evaluated using the Fundamental Theorem of Calculus, i.e. continuous functions on finite intervals. $\int_1^{\infty}{ \frac{dx}{x(x-2)} }$

Problem Statement

[10 points] For the following improper integral, set up but do not evaluate, limits and integrals such that the integrals can be evaluated using the Fundamental Theorem of Calculus, i.e. continuous functions on finite intervals. $\int_1^{\infty}{ \frac{dx}{x(x-2)} }$

$$\displaystyle{ \int_1^{\infty}{ \frac{dx}{x(x-2)} } = }$$ $$\displaystyle{ \lim_{b\to2^-}{\int_1^{b}{ \frac{dx}{x(x-2)} } } + }$$ $$\displaystyle{ \lim_{a\to2^+}{\int_a^{3}{ \frac{dx}{x(x-2)} } } + }$$ $$\displaystyle{ \lim_{c\to\infty}{\int_3^{c}{ \frac{dx}{x(x-2)} } } }$$

Problem Statement

[10 points] For the following improper integral, set up but do not evaluate, limits and integrals such that the integrals can be evaluated using the Fundamental Theorem of Calculus, i.e. continuous functions on finite intervals. $\int_1^{\infty}{ \frac{dx}{x(x-2)} }$

Solution

We need to check for the following discontinuities:
1. at each limit of integration
2. all discontinuities inside the integration interval
Lower limit: $$x=1$$ does not cause any issues since it is finite and is part of the domain of the integrand.
Upper limit: $$x=\infty$$ is an issue since we need a finite interval.
Now let's look at the integrand. For this problem, our only problems will occur when the denominator is zero. So we need to solve $$x(x-2)=0$$. This occurs when $$x=0$$ and $$x=2$$.
For $$x=0$$, this is outside the limits of integration, so it doesn't come into play here.
For $$x=2$$, this is inside our limits of integration, so we need to consider it.
So our only two issues are $$x=\infty$$ and $$x=2$$. First, let's break the integral at $$x=2$$ to isolate that point.

 $$\displaystyle{ \int_1^{2}{ \frac{dx}{x(x-2)} } + }$$ $$\displaystyle{ \int_2^{\infty}{ \frac{dx}{x(x-2)} } }$$

The first integral is ready to apply a limit since there is only one problem in the limits and none within the limits of integration. The second integral has two issues, one at each end. So we need to break it at a convenient point between 2 and $$\infty$$. We choose 3 but any point within the interval will work. So the second integral looks like this.

 $$\displaystyle{ \int_2^{3}{ \frac{dx}{x(x-2)} } + }$$ $$\displaystyle{ \int_3^{\infty}{ \frac{dx}{x(x-2)} } }$$

Now we are ready to combine our results and apply limits.
$$\displaystyle{ \lim_{b\to2^-}{\int_1^{b}{ \frac{dx}{x(x-2)} } } + }$$ $$\displaystyle{ \lim_{a\to2^+}{\int_a^{3}{ \frac{dx}{x(x-2)} } } + }$$ $$\displaystyle{ \lim_{c\to\infty}{\int_3^{c}{ \frac{dx}{x(x-2)} } } }$$

$$\displaystyle{ \int_1^{\infty}{ \frac{dx}{x(x-2)} } = }$$ $$\displaystyle{ \lim_{b\to2^-}{\int_1^{b}{ \frac{dx}{x(x-2)} } } + }$$ $$\displaystyle{ \lim_{a\to2^+}{\int_a^{3}{ \frac{dx}{x(x-2)} } } + }$$ $$\displaystyle{ \lim_{c\to\infty}{\int_3^{c}{ \frac{dx}{x(x-2)} } } }$$