Section Practice
¶These are practice problems for the tests. Solutions are in the next section. While we will not present these, I am happy to answer questions about them in class.
Here's a strategy for tackling indefinite integrals.
Is it easy? I.e. can we just guess a function \(f\) so that \(f'\) is the integrand?
Is there a function \(f\) in the integrand along with the derivative \(f'\) so that this is a reverse chain rule/subtitution problem, for example, \(\dsp \int x^2 \sin(4-x^3) \; dx\text{?}\)
Is there a substitution such as \(u = x+5\) for \(\dsp \int \frac{x}{\sqrt{x+5}} \; dx\text{,}\) or \(x = \tan(t)\) for \(\dsp \int \frac{x^2}{\sqrt{x^2+4}} \; dx?\)
Is there a way to rewrite the integrand using algebra, a trigonometric identity, long division, or partial fractions to make an easier problem?
Is the integrand of the form \(u \cdot v'\) for some functions \(u\) and \(v\) so that I can apply integration by parts?
Is there a power or a product of powers of trigonometric functions that I know to apply certain identities to?
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Is there some clever trick that I just know to apply to this integral such as when I see this particular problem? For example:
\(\dsp{ \int \cos^2(x) \; \; dx}\) Use the half-angle formula.
\(\dsp{ \int \sec(x) \; \; dx}\) Multiply by \(\dsp{ \frac{\sec(x) +\tan(x)} {\sec(x) + \tan(x)})}\text{.}\)
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Integrate by substitution.
\(\dsp \int x^2 \sqrt{x^3-3} \ dx\)
\(\dsp \int \frac{(\sqrt{x} + \pi)^4}{\sqrt{x}} \ dx\)
\(\dsp \int x\sqrt{x+5} \ dx\)
\(\dsp \int \sin(42x^5-6) 37x^4 \ dx\)
Evaluate \(\int x \sin(x^2) \cos(x^2) \; dx\) using \(u(x) = \sin(x^2).\) Now evaluate \(\int x \sin(x^2) \cos(x^2) \; dx\) using \(u(x) = \cos(x^2).\) Are your answers the same? Why or why not?
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Integrate by parts.
\(\dsp \int xe^{-x} \ dx\)
\(\dsp \int x\sin(2x) \ dx\)
\(\dsp \int x^2 \sinh(x) \ dx\) where \(\dsp \sinh(x) = \frac{e^x-e^{-x}}{2}\) and \(\dsp \cosh(x) = \frac{e^x+e^{-x}}{2}\)
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Integrate by partial fractions.
\(\dsp \int \frac{2}{x^2-1} \ dx\)
\(\dsp \int \frac{4x^2+13x-9}{x^3+2x^2-3x} \ dx\)
\(\dsp \int \frac{6x-11}{(x-1)^2} \ dx\)
\(\dsp \int \frac{-19x^2+50x-25}{x^2(3x-5)} \ dx\)
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Integrate by trigonometric identities.
\(\dsp \int \sin^3(x) \ dx\)
\(\dsp \int \cos^4(x) \ dx\)
\(\dsp \int \cos^3(x)\sin^6(x) \ dx\)
\(\dsp \int \tan^3(x)\sec^5(x) \ dx\)
\(\dsp \int \sin^2(x)\cos^2(x) \ dx\)
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Integrate by trigonometric substitution.
\(\dsp \int \frac{\sqrt{x^2-9}}{x} \ dx\)
\(\dsp \int \frac{1}{4+x^2} \ dx\) Either \(x=2\tan(t)\) or \(x = 2\sinh(t)\)) works.
\(\dsp \int \frac{x^2}{4-x^2} \ dx\)
\(\dsp \int \frac{1}{x^3\sqrt{x^2-25}} \ dx\)
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Integrate these Improper Integrals
\(\dsp \int_0^{\infty} {{3x} \over {4+x^2}}\; dx\)
\(\dsp \int_0^{\infty} {{7x} \over {x^4 + 9}}\; dx\)
\(\dsp \int_{-1}^{\infty} {{1} \over {x^2+2x+5}}\; dx\)
\(\dsp \int_0^1 {{-2} \over {x^3-5x^2}}\; dx\)
\(\dsp \int_{- \infty}^{\infty} {{2^x} \over {3+2^x}}\; dx\)
\(\dsp \int_0^e {{\ln(x)} \over {3x}}\; dx\)
Use the arc length formula to verify that the circumference the circle \(x^2+y^2=R^2\) is \(2\pi R\text{.}\) Where did you use the concept of improper integrals in this problem?
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Evaluate the indefinite integrals using any method.
\(\dsp \int \sqrt{x-7} \; dx \;\;\;\) \(\dsp \int x\sqrt{x-7} \; dx\; \; \;\) \(\dsp \int x^2\sqrt{x-7} \; dx\)
\(\dsp \int {{4t^2} \over {\sqrt{8-t^3}} } \; dt\)
\(\dsp \int{{3 \cot(t) + 2 \csc(t)} \over {\sin(t)} } \; dt\)
\(\dsp \int {{\sec^2 \sqrt{x+5}} \over {\sqrt{x+5}}} \; dx\)
\(\dsp \int_4^5 {{1} \over {\sqrt{x-4}}}\; dx\)
\(\dsp \int \cos (ax+b) \; dx\) Assume \(a\) and \(b\) are constants.
\(\dsp \int {3^{4x}} \; dx\)
\(\dsp \int {b^{kx}} \; dx\) Assume \(k\) is a constant.
\(\dsp \int (\sec(e^{2x}) + \csc(e^{2x}))^2 e^{2x} \; dx\)
\(\dsp \int{{2-e^{3x}} \over {e^x}}\; dx\)
\(\dsp \int{{e^{2x}} \over {e^x - 5}}\; dx\)
\(\dsp \int{{1} \over {3 \sqrt{x}(4+ \sqrt{x})^4}}\; dx\)
\(\dsp \int_0^{\infty} xe^{-2x}\; dx\)
\(\dsp \int {{\ln^3(4x)} \over {5x}}\; dx\)
\(\dsp \int x \sqrt{2x+1} \; dx\)
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Evaluate the indefinite integrals using any method.
\(\dsp \int {{2x-1} \over {x^3-5x^2-6x}}\; dx\)
\(\dsp \int \sin^3(x) \; dx\)
\(\dsp \int {{x^2-x-6} \over{x^3-1}}\;dx\)
\(\dsp \int (16-x^2)^{3/2}\; dx\)
\(\dsp \int {{3x+1} \over {x^2-4}}\; dx\)
\(\dsp \int \csc^6(x) \; dx\)
\(\dsp \int {{2x^3 -4x^2+x+ {{21} \over 8}} \over {2x^2-x-2}}\; dx\)
\(\dsp \int {{7x^2-3x-1} \over {x^3-x^2-x-2}}\; dx\)
\(\dsp \int \cos^2(x) \; dx\)
\(\dsp \int {4 \over {x \sqrt{x^4 + 9}}} \; dx\)
\(\dsp \int \sin^2(x) \cos^2(x) \; dx\)
\(\dsp \int \sin^3(x) \cos^4(x) \; dx\)
\(\dsp \int {{2x-1} \over {x^3 - 4x^2}}\; dx\)
\(\dsp \int \cot^5(x) \csc^2(x) \; dx\)
\(\dsp \int \tan^3(x)\; dx\)
\(\dsp \int \tan^4(x) \sec^4(x) \; dx\)
\(\dsp \int {2 \over {x^2 \sqrt{x^2 - 4}}}\; dx\)
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Evaluate the indefinite integrals using any method.
\(\dsp \int {3 \over {\sqrt{2x-x^2}}} \; dx\)
\(\dsp \int {{2x^3-4x^2+3x+1} \over {2x^2-x-2}} \; dx\)
\(\dsp \int {{7x^2-3x-1} \over {x^3-x^2-x-2}} \; dx\)
\(\dsp \int x \sin(3x) \; dx\)
\(\dsp \int x^2 4^x \; dx\)
\(\dsp \int x^5 \sqrt{x^2 + 4} \; dx\) Does \(u=\sqrt{x^2 + 4}\) or \(u = x^2+4\) work better?
\(\dsp \int {{\sqrt x} \over {1+ x^{1/3}}}\; dx\) (Let \(x = t^6\))
\(\dsp \int \log(x) \; dx\)
\(\dsp \int \invcos(x) \; dx\)
\(\dsp \int \sqrt{x^2-x-6} \; dx\)
\(\dsp \int x^3 \ln(x) \; dx\)
\(\dsp \int {6 \over {\sqrt{6x+x^2}}} \; dx\)
\(\dsp \int e^x \cos(2x) \; dx\)
\(\dsp \int \cos(x) \ln(\sin(x)) \; dx\)
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Assume that \(x\) is a number and show that each of the following is true.
\(1- \tanh^2(x) = \sech^2(x)\)
\(1- \coth^2(x) = -\csch^2(x)\)
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Assume that \(x\) is a number and show that each of the following is true.
\(\dsp \frac{d}{dx} \cosh(x) = \sinh(x)\)
\(\dsp \frac{d}{dx} \cotanh(x) = -\csch ^2(x)\)
\(\dsp \frac{d}{dx} \sech(x) = -\sech(x) \cdot \tanh(x)\)
\(\dsp \frac{d}{dx} \csch(x) = -\csch(x) \cdot \coth (x)\)
Rewrite \(\invsinh\) as a function of \(x\) and without using any hyperbolic functions by solving \(\dsp y = \frac{e^x-e^{-x}}{2}\) for \(x.\)
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Verify each of the following. Why are there restrictions on the domains?
\(\dsp \frac{d}{dx} \invcosh (x) = \frac{1}{\sqrt{x^2 -1}}\) for \(x > 1\)
\(\dsp \frac{d}{dx} \invtanh (x) = \frac{1}{1-x^2}\) for \(|x|\lt 1\)
\(\dsp \frac{d}{dx} \invcoth (x) = \frac{1}{1-x^2}\) for \(|x| > 1\)
\(\dsp \frac{d}{dx} \invcsch (x) = -\frac{1}{|x| \sqrt{1+x^2}}\) for \(x \ne 0\)