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Section Integration by Trigonometric Substitution

We have already used substitution (change of variables) to transform difficult integration problems into easier ones. Trigonometric substitutions are an extension of this same theme.

Problem 5.28.

Derive the formula (\(\dsp A=\pi r^2\)) for the area of a circle of radius \(r\) using integration as follows.

  1. Compute the definite integral \(\dsp \int_{-3}^3 \sqrt{9-x^2} \; dx\) by making the substitution, \(x(t) = 3\sin(t).\)

  2. Show that the area inside of a circle \(x^2+y^2=r^2\) is \(\pi r^2\) by setting up and evaluating an appropriate definite integral using the substitution \(x(t) = r\sin(t)\text{.}\)

Problem 5.29.

Evaluate each of the following integrals.

  1. \(\dsp \int \frac{x^2}{\sqrt{x^2+4}}\; dx \;\) by using the substitution \(x = 2 \tan (\theta)\)

  2. \(\dsp \int_0^4 5x \sqrt{16-x^2} \; dx\)

  3. \(\dsp \int_2^3 \frac{1}{{4x^3 \sqrt{x^2 - 1}}} \; dx \;\) by using the substitution \(x = \sec (\theta)\)

  4. \(\dsp \int \frac{1}{\sqrt{x^2-2x + 5}} \; dx \;\) by completing square and letting \(u=x-1\)

  5. \(\dsp \int \frac{2x+5}{x^2+4x+1} \; dx \;\)

Definition 5.30.

If \(a\) and \(b\) are non-zero numbers, then the set of points in the plane satisfying

\begin{equation*} \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \end{equation*}

is an ellipse centered at the origin.

Problem 5.31.

Consider the ellipse, \(\dsp \frac{x^2}{25} + \frac{y^2}{16} = 1\text{.}\) Set up and evaluate a definite integral that represents the area (or half the area) inside this ellipse.

Problem 5.32.

Solve each of the following problems. Assume that \(a\) is a real number.

  1. Evaluate \(\dsp \int \frac{1}{\sqrt{a^2 - x^2}} \; dx\)

  2. Evaluate \(\dsp \int \frac{1}{x^2 + a^2} \; dx\)

  3. Evaluate \(\dsp \int \frac{1}{x\sqrt{x^2 - a^2}} \; dx\)

Problem 5.33.

Verify the formula \(\dsp A=\pi ab\) for the area of the ellipse \(\dsp \frac{x^2}{a^2}+\frac{y^2}{b^2} = 1\) where \(a\) and \(b\) are posiive numbers.

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