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Section 2.3 Trigonometric Substitutions

So far we have seen that it sometimes helps to replace a subexpression of a function by a single variable. Occasionally it can help to replace the original variable by something more complicated. This seems like a “reverse” substitution, but it is really no different in principle than ordinary substitution.

Example 2.26. Sine Substitution.

Evaluate \(\ds\int \sqrt{1-x^2}\,dx\text{.}\)


Let \(x=\sin u\) so \(dx=\cos u\,du\text{.}\) Then

\begin{equation*} \int \sqrt{1-x^2}\,dx=\int\sqrt{1-\sin^2 u}\cos u\,du= \int\sqrt{\cos^2 u}\cos u\,du\text{.} \end{equation*}

We would like to replace \(\ds \sqrt{\cos^2 u}\) by \(\cos u\text{,}\) but this is valid only if \(\cos u\) is positive, since \(\ds \sqrt{\cos^2 u}\) is positive. Consider again the substitution \(x=\sin u\text{.}\) We could just as well think of this as \(u=\arcsin x\text{.}\) If we do, then by the definition of the arcsine, \(-\pi/2\le u\le\pi/2\text{,}\) so \(\cos u\ge0\) and so we are allowed to continue and perform the simplification:

\begin{align*} \int\sqrt{\cos^2 u}\cos u\,du =\amp \int\cos^2u\,du\\ =\amp \int {1+\cos 2u\over2}\,du = {u\over 2}+{\sin 2u\over4}+C\\ =\amp {\arcsin x\over2}+{\sin(2\arcsin x)\over4}+C \end{align*}

This is a perfectly good answer, though the term \(\sin(2\arcsin x)\) is a bit unpleasant. It is possible to simplify this. Using the identity \(\sin 2x=2\sin x\cos x\text{,}\) we can write

\begin{equation*} \begin{split} \ds \sin 2u=\amp2\sin u\cos u=2\sin(\arcsin x)\sqrt{1-\sin^2 u}\\ \amp= 2x\sqrt{1-\sin^2(\arcsin x)}=2x\sqrt{1-x^2}\end{split}\text{.} \end{equation*}

Then the full antiderivative is

\begin{equation*} {\arcsin x\over2}+{2x\sqrt{1-x^2}\over4}+C= {\arcsin x\over2}+{x\sqrt{1-x^2}\over2}+C\text{.} \end{equation*}


  1. This type of substitution is usually indicated when the function you wish to integrate contains a polynomial expression that might allow you to use the fundamental identity \(\ds \sin^2x+\cos^2x=1\) in one of three forms:

    \begin{equation*} \cos^2 x=1-\sin^2x \qquad \sec^2x=1+\tan^2x \qquad \tan^2x=\sec^2x-1\text{.} \end{equation*}

    If your function contains \(\ds 1-x^2\text{,}\) as in the example above, try \(x=\sin u\text{;}\) if it contains \(\ds 1+x^2\) try \(x=\tan u\text{;}\) and if it contains \(\ds x^2-1\text{,}\) try \(x=\sec u\text{.}\) Sometimes you will need to try something a bit different to handle constants other than inverse substitution, which is described next.

  2. In a traditional substitution we let \(u=u(x)\text{,}\) i.e., our new variable is defined in terms of \(x\text{.}\) In an inverse substitution we let \(x=g(u)\text{,}\) i.e., we assume \(x\) can be written in terms of \(u\text{.}\) We cannot do this arbitrarily since we do NOT get to “choose” \(x\text{.}\) For example, an inverse substitution of \(x=1\) will give an obviously wrong answer. However, when \(x=g(u)\) is an invertible function, then we are really doing a \(u\)-substitution with \(u=g^{-1}(x)\text{.}\) Now the Substitution Rule applies.

  3. Sometimes with inverse substitutions involving trig functions we use \(\theta\) instead of \(u\text{.}\) Thus, we would take \(x=\sin\theta\) instead of \(x=\sin u\text{.}\)

  4. We would like our inverse substitution \(x=g(u)\) to be a one-to-one function, and \(x=\sin u\) is not one-to-one. In the next few paragraphs, we discuss how we can overcome this issue by using the restricted trigonometric functions.

The three common trigonometric substitutions are the restricted sine, restricted tangent and restricted secant. Thus, for sine we use the domain \([-\pi/2,~\pi/2]\) and for tangent we use \((-\pi/2,~\pi/2)\text{.}\) Depending on the convention chosen, the restricted secant function is usually defined in one of two ways.

One convention is to restrict secant to the region \([0,~\pi/2)\cup(\pi/2,\pi]\) as shown in the middle graph. The other convention is to use \([0,~\pi/2)\cup[\pi,~3\pi/2)\) as shown in the right graph. Both choices give a one-to-one restricted secant function and no universal convention has been adopted. To make the analysis in this section less cumbersome, we will use the domain \([0,~\pi/2)\cup[\pi,~3\pi/2)\) for the restricted secant function. Then \(\sec^{-1}x\) is defined to be the inverse of this restricted secant function.

Typically trigonometric substitutions are used for problems that involve radical expressions. The table below outlines when each substitution is typically used along with their restricted intervals.

\begin{equation*} \begin{array}{ccc} \mbox{Expression} \amp \mbox{Substitution} \amp \mbox{Restricted Interval} \\ \hline\\ \sqrt{a^2-x^2} \amp x=a\sin\theta \amp \theta\in[-\pi/2,~\pi/2]\\ ~ \amp ~ \amp ~ \\~ \amp ~ \amp ~ \\ \sqrt{a^2+x^2}~~\mbox{or} ~~a^2+x^2 \amp x=a\tan\theta \amp \theta\in(-\pi/2,~\pi/2)\\ ~ \amp ~ \amp ~ \\~ \amp ~ \amp ~ \\ \sqrt{x^2-a^2} \amp x=a\sec\theta \amp \theta\in[0,~\pi/2)\cup[\pi,~3\pi/2) \end{array} \end{equation*}

}All three substitutions are one-to-one on the listed intervals. When dealing with radicals we often end up with absolute values since

\begin{equation*} \sqrt{z^2}=|z|\text{.} \end{equation*}

For each of the three trigonometric substitutions above we will verify that we can ignore the absolute value in each case when encountering a radical.

For \(x=a\sin\theta\text{,}\) the expression \(\sqrt{a^2-x^2}\) becomes

\begin{equation*} \sqrt{a^2-x^2} = \sqrt{a^2-a^2\sin^2\theta} = \sqrt{a^2(1-\sin^2\theta)} = a\sqrt{\cos^2\theta} = a|\cos\theta| = a\cos\theta \end{equation*}

This is because \(\cos\theta\geq 0\) when \(\theta\in[-\pi/2,~\pi/2]\text{.}\) For \(x=a\tan\theta\text{,}\) the expression \(\sqrt{a^2+x^2}\) becomes

\begin{equation*} \sqrt{a^2+x^2} = \sqrt{a^2+a^2\tan^2\theta} = \sqrt{a^2(1+\tan^2\theta)} = a\sqrt{\sec^2\theta} = a|\sec\theta| = a\sec\theta \end{equation*}

This is because \(\sec\theta>0\) when \(\theta\in(-\pi/2,~\pi/2)\text{.}\)

Finally, for \(x=a\sec\theta\text{,}\) the expression \(\sqrt{x^2-a^2}\) becomes

\begin{equation*} \sqrt{x^2-a^2} = \sqrt{a^2\sec^2\theta-a^2} = \sqrt{a^2(\sec^2\theta-1)} = a\sqrt{\tan^2\theta} = a|\tan\theta| = a\tan\theta \end{equation*}

This is because \(\tan\theta\geq 0\) when \(\theta\in[0,~\pi/2)\cup[\pi,~3\pi/2)\text{.}\)

Thus, when using an appropriate trigonometric substitution we can usually ignore the absolute value. After integrating, we typically get an answer in terms of \(\theta\) (or \(u\)) and need to convert back to \(x\)'s. To do so, we use the guideline below:

  • For trig functions containing \(\theta\text{,}\) use a triangle to convert to \(x\)'s.

  • For \(\theta\) by itself, use the inverse trig function.

All pieces needed for such a trigonometric substitution can be summarized as follows:

Guideline for Trigonometric Substitution.

Suppose we have an integral with any of the following expressions, then use the substitution, differential, identity and inverse of substitution listed below to guide yourself through the integration process:

\begin{equation*} \begin{array}{ccccc} \amp \amp \amp \amp \textbf{Inverse of} \\ \textbf{Expression} \amp \textbf{Substitution} \amp \textbf{Differential} \amp \textbf{Identity} \amp \textbf{Substitution} \\ \hline ~ \amp ~ \amp ~ \amp ~ \amp ~ \\ \sqrt{a^2-x^2} \amp x=a\sin\theta \amp dx=a\cos\theta d\theta \amp \sqrt{a^2-x^2}=a\cos\theta \amp \theta=\sin^{-1}\left(\frac{x}{a}\right) \\ ~ \amp ~ \amp ~ \amp ~ \amp ~ \\ ~ \amp ~ \amp ~ \amp ~ \amp ~ \\ \sqrt{a^2+x^2} \amp \amp \amp \amp \\ \text{ or } \amp x=a\tan\theta \amp dx=a\sec^2\theta d\theta \amp \sqrt{a^2+x^2}=a\sec\theta \amp \theta=\tan^{-1}\left(\frac{x}{a}\right) \\ a^2+x^2 \amp \amp \amp \amp \\ ~ \amp ~ \amp ~ \amp ~ \amp ~ \\ ~ \amp ~ \amp ~ \amp ~ \amp ~ \\ \sqrt{x^2-a^2} \amp x=a\sec\theta \amp dx=a\sec\theta\tan\theta d\theta \amp \sqrt{x^2-a^2}=a\tan\theta \amp \theta=\sec^{-1}\left(\frac{x}{a}\right) \end{array} \end{equation*}

To emphasize the technique, we redo the computation for \(\ds\int \sqrt{1-x^2}\,dx\text{.}\)

Example 2.27. Sine Substitution.

Evaluate \(\ds\int \sqrt{1-x^2}\,dx\text{.}\)


Since \(\sqrt{1-x^2}\) appears in the integrand we try the trigonometric substitution \(x=\sin\theta\text{.}\) (Here we are using the restricted sine function with \(\theta\in[-\pi/2,~\pi/2]\) but typically omit this detail when writing out the solution.) Then \(dx=\cos \theta\,d\theta\text{.}\)

\begin{align*} \int \sqrt{1-x^2}\,dx = \amp \int\sqrt{1-\sin^2\theta}\,\cos \theta\,d\theta \amp\amp \mbox{Using our (inverse) substitution}\\ = \amp \int \sqrt{\cos^2\theta}\cos \theta\,d\theta \amp\amp \mbox{Since \(\sin^2\theta+\cos^2\theta=1\)} \\ = \amp \int |\cos\theta|\cdot\cos\theta\,d\theta \amp \amp \mbox{Since \(\sqrt{\cos^2\theta}=|\cos\theta|\)} \\ = \amp \int \cos^2\theta\,d\theta, \amp\amp \mbox{ } \end{align*}

where, in the last line, we had \(|\cos \theta| = \cos \theta| \) since for \(\theta\in[-\frac{\pi}{2},\frac{\pi}{2}]\) we have \(\cos\theta\geq0\text{.}\)

Often we omit the step containing the absolute value by our discussion above. Now, to integrate a power of cosine we use the guideline for products of sine and cosine and make use of the identity

\begin{equation*} \cos^2\theta=\frac{1}{2}(1+\cos(2\theta))\text{.} \end{equation*}

Our integral then becomes

\begin{equation*} \int \sqrt{1-x^2}\,dx=\frac{1}{2}\int (1+\cos(2\theta))\,d\theta=\frac{\theta}{2}+\frac{\sin(2\theta)}{4}+C \end{equation*}

To write the answer back in terms of \(x\) we use a right triangle. Since \(\sin\theta=x/1\) we have the triangle:

The triangle gives \(\sin\theta\text{,}\) \(\cos\theta\text{,}\) \(\tan\theta\text{,}\) but have a \(\sin(2\theta)\text{.}\) Thus, we use an identity to write

\begin{equation*} \sin(2\theta) = 2\sin\theta\cos\theta = 2\left(\frac{x}{1}\right)\left(\frac{\sqrt{1-x^2}}{1}\right) \end{equation*}

For \(\theta\) by itself we use \(\theta=\sin^{-1}x\text{.}\) Thus, the integral is

\begin{equation*} \int \sqrt{1-x^2}\,dx =\frac{\sin^{-1}x}{2}+\frac{x\sqrt{1-x^2}}{2}+C \end{equation*}
Example 2.28. Secant Substitution.

Evaluate \(\ds\int\frac{\sqrt{25x^2-4}}{x}\,dx\text{.}\)


We do not have \(\sqrt{x^2-a^2}\) because of the \(25\text{,}\) but if we factor \(25\) out we get:

\begin{equation*} \ds\int\frac{\sqrt{25(x^2-(4/25))}}{x}\,dx=\ds\int5\frac{\sqrt{x^2-(4/25)}}{x}\,dx\text{.} \end{equation*}

Now, \(a=2/5\text{,}\) so let \(x=\frac{2}{5}\sec\theta\text{.}\) Alternatively, we can think of the integral as being:

\begin{equation*} \ds\int\frac{\sqrt{(5x)^2-4}}{x}\,dx \end{equation*}

Then we could let \(u=5x\) followed by \(u=2\sec\theta\text{,}\) etc. Or equivalently, we can avoid a \(u\)-substitution by letting \(5x=2\sec\theta\text{.}\) In either case we are using the trigonometric substitution \(x=\frac{2}{5}\sec\theta\text{,}\) but do use the method that makes the most sense to you! As \(x=\frac{2}{5}\sec\theta\) we have \(dx=\frac{2}{5}\sec\theta\tan\theta\,d\theta\text{.}\)

\begin{align*} \int\frac{\sqrt{25x^2-4}}{x}\,dx = \amp \int \frac{\sqrt{25\frac{4\sec^2\theta}{25}-4}}{\frac{2}{5}\sec\theta}\,\frac{2}{5}\sec\theta\tan\theta\,d\theta \amp \amp\mbox{Using the substitution} \\ = \amp \int\sqrt{4(\sec^2\theta-1)}\cdot \tan\theta\,d\theta \amp\amp \mbox{Cancelling} \\ = \amp 2\int\sqrt{\tan^2\theta}\cdot \tan\theta\,d\theta \amp\amp \mbox{Using \(\tan^2\theta+1=\sec^2\theta\)} \\ = \amp 2\int\tan^2\theta\,d\theta \amp\amp \mbox{Simplifying} \\ = \amp 2\int(\sec^2\theta-1)\,d\theta \amp\amp \mbox{Using \(\tan^2\theta+1=\sec^2\theta\)} \\ = \amp 2(\tan\theta-\theta)+C \amp\amp \mbox{Since \(\ds\int\sec^2\theta\,d\theta=\tan\theta+C\)} \end{align*}

For \(\tan\theta\text{,}\) we use a right triangle.

\begin{equation*} x=\frac{2}{5}\sec\theta\qquad\to \qquad x=\frac{2}{5}\frac{1}{\cos\theta}\qquad\to \qquad \cos\theta=\frac{2}{5x} \end{equation*}

Using SOH CAH TOA, the triangle is then

For \(\theta\) by itself, we use \(\theta=\sec^{-1}(5x/2)\text{.}\) Thus,

\begin{equation*} \begin{split} \ds\int\frac{\sqrt{25x^2-4}}{x}\,dx=\amp 2\left(\frac{\sqrt{25x^2-4}}{2}-\sec^{-1}\left(\frac{5x}{2}\right)\right)+C\\ =\amp \sqrt{25x^2-4}-2\sec^{-1}\left(\frac{5x}{2}\right)+C. \end{split} \end{equation*}

In the context of the previous example, some resources give an alternate guideline when choosing a trigonometric substitution.

\begin{equation*} \begin{split} \sqrt{a^2-b^2x^2} \amp \to x=\ds\frac{a}{b}\sin\theta\\ \sqrt{b^2x^2+a^2}~~\mbox{or} ~~(b^2x^2+a^2) \amp \to x=\ds\frac{a}{b}\tan\theta\\ \sqrt{b^2x^2-a^2}\amp \to x=\ds\frac{a}{b}\sec\theta\end{split} \end{equation*}

We next look at a tangent substitution.

Example 2.29. Tangent Substitution.

Evaluate \(\ds\int\frac{1}{\sqrt{25+x^2}}\,dx\text{.}\)


Let \(x=5\tan\theta\) so that \(dx=5\sec^2\theta\,d\theta\text{.}\)

\begin{align*} \int\frac{1}{\sqrt{25+x^2}}\,dx = \amp \int \frac{1}{\sqrt{25+25\tan^2\theta}}\,5\sec^2\theta\,d\theta \amp\amp \mbox{Using our substitution}\\ = \amp \int\frac{1}{\sqrt{25(1+\tan^2\theta)}}\cdot5\sec^2\theta\,d\theta \amp\amp \mbox{Factor out \(25\)} \\ = \amp \int\frac{1}{5\sqrt{sec^2\theta}}\cdot5\sec^2\theta\,d\theta \amp\amp \mbox{Using \(\tan^2\theta+1=\sec^2\theta\)} \\ = \amp \int\sec\theta\,d\theta \amp\amp \mbox{Simplifying} \\ = \amp \ln|\sec\theta+\tan\theta|+C \amp\amp \mbox{By \(\ds\int\sec \theta\,dx=\ln|\sec \theta+\tan \theta|+C\)} \end{align*}

Since \(\tan\theta=x/5\text{,}\) we draw a triangle:


\begin{equation*} \ds\sec\theta=\frac{1}{\cos\theta}=\frac{\sqrt{25+x^2}}{5}\text{.} \end{equation*}

Therefore, the integral is

\begin{equation*} \int\frac{1}{\sqrt{25+x^2}}\,dx=\ln\left|\frac{\sqrt{25+x^2}}{5}+\frac{x}{5}\right|+C \end{equation*}

In the next example, we will use the technique of completing the square in order to rewrite the integrand.

Example 2.30. Completing the Square.

Evaluate \(\ds\int\frac{x}{\sqrt{3-2x-x^2}}\,dx\text{.}\)


First, complete the square to write

\begin{equation*} 3-2x-x^2=4-(x+1)^2 \end{equation*}

Now, we may let \(u=x+1\) so that \(du=dx\) (note that \(x=u-1\)) to get:

\begin{equation*} \int\frac{x}{\sqrt{4-(x+1)^2}}\,dx=\int\frac{u-1}{\sqrt{4-u^2}}\,du \end{equation*}

Let \(u=2\sin\theta\) giving \(du=2\cos\theta\,d\theta\text{:}\)

\begin{equation*} \int\frac{u-1}{\sqrt{4-u^2}}\,du = \int\frac{2\sin\theta-1}{2\cos\theta}\cdot 2\cos\theta\,d\theta=\int (2\sin\theta-1)\,d\theta \end{equation*}

Integrating and using a triangle we get:

\begin{align*} \int\frac{x}{\sqrt{3-2x-x^2}} = \amp -2\cos\theta-\theta+C\\ = \amp -\sqrt{4-u^2}-\sin^{-1}\left(\frac{u}{2}\right)+C\\ = \amp -\sqrt{3-2x-x^2}-\sin^{-1}\left(\frac{x+1}{2}\right)+C \end{align*}

Note that in this problem we could have skipped the \(u\)-substitution if instead we let \(x+1=2\sin\theta\text{.}\) (For the triangle we would then use \(\ds\sin\theta=\frac{x+1}{2}\text{.}\))

Exercises for Section 2.3.

Evaluate the following indefinite integrals.

  1. \(\ds\int\sqrt{x^2-1}\,dx\)

    \(\ds x\sqrt{x^2-1}/2-\ln|x+\sqrt{x^2-1}|/2+C\)

    We use the substitution:

    \begin{equation*} x = \sec\theta,\ dx = \sec\theta\tan\theta\,d\theta\text{.} \end{equation*}


    \begin{equation*} \begin{split} \int \sqrt{x^2-1} \,dx \amp = \int \sqrt{\sec^2\theta-1}\,\sec\theta\tan\theta\,d\theta\\ \amp = \int \tan^2\theta \sec\theta\,d\theta\\ \amp = \int \sec^3\theta - \sec\theta \,d\theta\\ \amp = \frac{1}{2} \sec\theta\tan\theta + \frac{1}{2}\ln|\tan\theta+\sec\theta|-\ln|\sec\theta + \tan\theta+C\\ \amp = \frac{1}{2} \sec\theta\tan\theta -\frac{1}{2}\ln|\tan\theta+\sec\theta|+C. \end{split} \end{equation*}

    We now need to write the solution in terms of \(x\text{.}\) Since \(\sec\theta = x\text{,}\) this means that \(\tan\theta = \sqrt{x^2-1}\text{.}\) Thus:

    \begin{equation*} \int \sqrt{x^2-1} = \frac{1}{2}x\sqrt{x^2-1} -\frac{1}{2}\ln|\sqrt{x^2-1}+x|+C\text{.} \end{equation*}
  2. \(\ds\int\sqrt{9+4x^2}\,dx\)

    \(\ds x\sqrt{9+4x^2}/2+\hbox{\)\ds(9/4)\ln|2x+\sqrt{9+4x^2}|+C\(}\)

    We first rewrite the integrand as

    \begin{equation*} \sqrt{9+4x^2} = 2\sqrt{\frac{9}{4}+x^2}\text{.} \end{equation*}

    Now let

    \begin{equation*} x=\frac{3}{2}\tan\theta,\ dx = \frac{3}{2}\sec^2\theta\,d\theta\text{.} \end{equation*}

    Then the integral becomes

    \begin{equation*} \int 2\sqrt{\frac{9}{4}+x^2} \,dx = 2\int \frac{9}{4}\sec^3\theta\,d\theta\text{.} \end{equation*}

    Therefore, we find that

    \begin{equation*} \int \sqrt{9+4x^2}\,dx = \frac{9}{4} \sec\theta\tan\theta + \frac{9}{4}\ln|\tan\theta+\sec\theta|+C\text{.} \end{equation*}

    If \(x= \frac{3}{2}\tan\theta\text{,}\) then we must have \(\sec\theta = \sqrt{1+4x^2/9}\text{.}\) Therefore, we find that

    \begin{equation*} \int \sqrt{9+4x^2}\,dx = \frac{2x}{3}\sqrt{4x^2/9+1} + \frac{9}{4}\ln\left(\sqrt{\frac{4x^2}{9}+1}+\frac{2x}{3}\right)+C \end{equation*}
  3. \(\ds\int x\sqrt{1-x^2}\,dx\)

    \(\ds -(1-x^2)^{3/2}/3+C\)

    Let \(u=1-x^2\) with \(du = -2x\,dx\text{.}\) Then:

    \begin{equation*} \begin{split} \int x\sqrt{1-x^2}\,dx \amp = \frac{-1}{2}\int \sqrt{u}\,du\\ \amp = \frac{-1}{3} u^{3/2} + C\\ \amp = \frac{-1}{3} \left(1-x^2\right)^{3/2}. \end{split} \end{equation*}
  4. \(\ds\int x^2\sqrt{1-x^2}\,dx\)

    \(\arcsin(x)/8-\sin(4\arcsin x)/32+C\)

    We see that \(\sqrt{1-x^2}\) is in the integrand, and so try the substitution:

    \begin{equation*} x = \sin\theta, \ dx = -\sin\theta\, d\theta \text{ for } \theta \in [-\pi/2,\pi/2]\text{.} \end{equation*}

    This gives

    \begin{equation*} \begin{split} \int x^2\sqrt{1-x^2} \amp = \int \sin^2\theta \sqrt{\cos^2\theta} \cos\theta\, d\theta = \int \sin^2\theta |\cos\theta| \cos\theta\, d\theta \\ \amp = \int \sin^2\theta \cos^2\theta \, d\theta \end{split} \end{equation*}

    Now use the identities \(\cos^2\theta = \frac{1}{2} (1+\cos(2\theta))\) and \(\sin^2\theta = \frac{1}{2}(1-\cos(2\theta))\text{:}\)

    \begin{equation*} \begin{split} \int \sin^2\theta \cos^2\theta \, d\theta \amp = \int \frac{1}{4}\left[(1-\cos (2\theta))(1+\cos(2\theta))\right] \, d\theta \\ \amp = \int \frac{1}{4} \left[ 1- \cos(2\theta) + \cos(2\theta) - \cos^2 (2\theta)\right]\, d\theta \\ \amp = \int \frac{1}{4} \left[ 1 - \frac{1}{2}\left(1+\cos(4\theta)\right)\right]\, d\theta \\ \amp = \frac{1}{8} \theta - \frac{1}{32} \sin(4\theta) + C \end{split} \end{equation*}

    Rewriting in terms of \(x\text{,}\) we find

    \begin{equation*} \int x^2\sqrt{1-x^2}\, dx = \frac{1}{8} \arcsin x - \frac{1}{32} \sin (4\arcsin x) + \hat{C}\text{.} \end{equation*}
  5. \(\ds\int{1\over\sqrt{1+x^2}}\,dx\)

    \(\ds \ln|x+\sqrt{1+x^2}|+C\)

    We take \(x=\tan\theta\text{,}\) with \(dx = \sec^2\theta d\theta\text{.}\) Therefore,

    \begin{equation*} \begin{split} \int \frac{1}{\sqrt{1+x^2}}\,dx \amp = \int \frac{\sec^2\theta}{\sqrt{1+\tan^2\theta}} \, d\theta \\ \amp = \int \frac{\sec^2\theta}{\sqrt{\sec^2\theta}}d\theta \\ \amp = \int \sec\theta d\theta = \ln \lvert \sec\theta + \tan\theta \rvert + C. \end{split} \end{equation*}

    We know that \(\tan \theta = x\text{,}\) and so \(\sec \theta= \sqrt{x^2+1}\text{.}\) Therefore,

    \begin{equation*} \int \frac{1}{\sqrt{1+x^2}}\,dx = \ln \lvert \sqrt{x^2+1} + x\rvert + \hat{C}\text{.} \end{equation*}
  6. \(\ds\int\sqrt{x^2+2x}\,dx\)

    \(\ds (x+1)\sqrt{x^2+2x}/2-\hbox{\)\ds\ln|x+1+\sqrt{x^2+2x}|/2+C\(}\)

    First notice that \(x^2+2x = (x+1)^2-1\text{.}\) So we first make the substitution,

    \begin{equation*} u = x+1, \ du = dx \end{equation*}


    \begin{equation*} \int \sqrt{x^2+2x} \, dx = \int \sqrt{(x+1)^2 - 1} \, dx = \int \sqrt{u^2-1} \, du\text{.} \end{equation*}

    We now make a secant substitution. Let \(u = \sec \theta\) with \(du = \sec\theta \tan\theta \,d\theta\text{:}\)

    \begin{equation*} \begin{split} \int \sqrt{u^2-1} \, du \amp = \int \sqrt{\sec^2\theta - 1} \sec\theta \tan\theta \, d\theta \\ \amp = \int \tan^2\theta \sec\theta \, d\theta \\ \amp = \int\sec\theta (\sec^2\theta - 1) \, d\theta \\ \amp = \int (\sec^3\theta - \sec\theta) d\theta \\ \amp = \frac{1}{2} \sec\theta \tan\theta - \frac{1}{2} \ln |\sec\theta + \tan\theta| + C. \end{split} \end{equation*}

    Since \(\sec \theta=u\text{,}\) we must have that \(\tan\theta = \sqrt{u^2-1}\text{.}\) Therefore,

    \begin{equation*} \begin{split} \int \sqrt{x^2+2x} \, dx \amp = \frac{1}{2} u \sqrt{u^2-1} - \frac{1}{2} \ln |u + \sqrt{u^2-1}| + \hat{C}\\ \amp = \frac{1}{2} (x-1) \sqrt{(x-1)^2-1} - \frac{1}{2} \ln |(x-1) + \sqrt{(x-1)^2-1}| + \overline{C}, \end{split} \end{equation*}

    where we used \(x=u-1\text{.}\)

  7. \(\ds\int{1\over x^2(1+x^2)}\,dx\)

    \(-\arctan x - 1/x+C\)

    We use the following substitution:

    \begin{equation*} x = \tan\theta \text{ with } dx = \sec^2\theta\,d\theta \text{.} \end{equation*}

    Therefore, we have

    \begin{equation*} \begin{split} \int \frac{1}{x^2(1+x^2)} \, dx \amp = \int \frac{1}{\tan^2\theta (1+\tan^2\theta) }\sec^2\theta\,d\theta \\ \amp = \int \frac{1}{\tan^2\theta \sec^2\theta }\sec^2\theta\,d\theta\\ \amp = \int \frac{1}{\tan^2\theta}\,d\theta\\ \amp = \int \cot^2 \theta \,d\theta \end{split} \end{equation*}

    We now use the identity \(\cot^2\theta = \csc^2\theta-1\text{:}\)

    \begin{equation*} \begin{split} \int \cot^2 \theta \,d\theta \amp = \int \csc^2\theta \,d\theta - \int \,d\theta\\ \amp = -\cot\theta - \theta + C. \end{split} \end{equation*}

    Since \(x=\tan\theta\text{,}\) we have

    \begin{equation*} \theta = \arctan x\text{,} \end{equation*}


    \begin{equation*} \cot\theta = \frac{1}{x}\text{.} \end{equation*}

    We find that

    \begin{equation*} \int \frac{1}{x^2(1+x^2)} \, dx = -\frac{1}{x} - \arctan x + C\text{.} \end{equation*}
  8. \(\ds\int{x^2\over\sqrt{4-x^2}}\,dx\)

    \(\ds 2\arcsin(x/2)-x\sqrt{4-x^2}/2+C\)


    \begin{equation*} x = 2\sin\theta, \text{ with } dx = 2\cos\theta\,d\theta\text{.} \end{equation*}

    Then the integral becomes

    \begin{equation*} \begin{split} \int \frac{x^2}{\sqrt{4-x^2}} \, dx \amp = \int \frac{4 \sin^2\theta}{\sqrt{4-4\sin^2\theta}} \left(2 \cos\theta\right)\,d\theta\\ \amp = \int \frac{4 \sin^2\theta}{2\cos\theta} \left(2 \cos\theta\right)\,d\theta\\ \amp =4\int \sin^2\theta\,d\theta \end{split} \end{equation*}

    Now to evaluate this integral, we let

    \begin{equation*} \sin^2 \theta = \frac{1}{2}(1-\cos(2\theta))\text{.} \end{equation*}


    \begin{equation*} 4\int \sin^2 \theta \,d\theta = 2 \int 1-\cos (2\theta) \,d\theta\text{.} \end{equation*}

    Now let \(u=2\theta\) with \(du = 2\,d\theta\text{:}\)

    \begin{equation*} \begin{split} 2 \int 1-\cos (2\theta) \,d\theta \amp = \int (1-\cos(u))\,du\\ \amp = u - \sin u + C \end{split} \end{equation*}


    \begin{equation*} 4 \int \sin^2 \theta \,d\theta = 2\theta - \sin (2\theta) + C\text{.} \end{equation*}

    Now since \(x = 2\sin\theta\text{,}\) we have

    \begin{equation*} \theta = \arcsin\left(\frac{\theta}{2}\right)\text{,} \end{equation*}


    \begin{equation*} \sin (2\theta) = 2\sin\theta\cos\theta = x \frac{\sqrt{1-x^2}}{2}\text{.} \end{equation*}

    We have

    \begin{equation*} \int \frac{x^2}{\sqrt{4-x^2}} \, dx= 2\arcsin\left(\frac{\theta}{2}\right) - \frac{x\sqrt{1-x^2}}{2} + C\text{.} \end{equation*}
  9. \(\ds\int{\sqrt{x}\over\sqrt{1-x}}\,dx\)

    \(\ds \arcsin(\sqrt{x})-\sqrt{x}\sqrt{1-x}+C\)

    We try the following substitution:

    \begin{equation*} \sqrt{x} = \sin \theta \implies x = \sin^2\theta\text{,} \end{equation*}


    \begin{equation*} dx = 2\sin\theta\cos\theta\,d\theta\text{.} \end{equation*}

    Then the integral becomes

    \begin{equation*} \begin{split} \int \frac{\sqrt{x}}{\sqrt{1-x}} \, dx \amp = \int \frac{\sin\theta}{\sqrt{1-\sin^2\theta}}2\sin\theta\cos\theta \,d\theta\\ \amp = \int \frac{\sin\theta}{\cos \theta} 2 \sin\theta\cos\theta\,d\theta\\ \amp = 2 \int \sin^2\theta\,d\theta. \end{split} \end{equation*}

    Now let

    \begin{equation*} 2 \sin^2 \theta = \frac{1}{2}(1-\cos(2\theta))\text{.} \end{equation*}


    \begin{equation*} 2\int \sin^2 \theta \,d\theta = \int 1-\cos (2\theta) \,d\theta\text{.} \end{equation*}

    Now let \(u=2\theta\) with \(du = 2\,d\theta\text{:}\)

    \begin{equation*} \begin{split} \int 1-\cos (2\theta) \,d\theta \amp = \frac{1}{4} \int (1-\cos(u))\,du\\ \amp =\frac{1}{2}\left( u - \sin u\right) + C \end{split} \end{equation*}


    \begin{equation*} 2\int \sin^2 \theta \,d\theta = \theta - \frac{1}{2} \sin (2\theta) + C\text{.} \end{equation*}

    Recall that \(\sin(2\theta) = 2\sin\theta\cos\theta\text{.}\) Therefore:

    \begin{equation*} 2\int \sin^2 \theta \,d\theta = \theta -\sin\theta\cos\theta + C\text{.} \end{equation*}

    Now we use the fact that

    \begin{equation*} \theta = \arcsin \sqrt{x}\text{,} \end{equation*}


    \begin{equation*} \sin\theta\cos\theta = \sqrt{x} \sqrt{1-x} \end{equation*}

    to write

    \begin{equation*} \int \frac{\sqrt{x}}{\sqrt{1-x}} \, dx = \arcsin\sqrt{x}- \sqrt{x}\sqrt{1-x} + C\text{.} \end{equation*}
  10. \(\ds\int{x^3\over\sqrt{4x^2-1}}\,dx\)

    \(\ds (2x^2+1)\sqrt{4x^2-1}/24+C\)

    We first rewrite the integrand:

    \begin{equation*} \frac{x^3}{\sqrt{4x^2-1}} = \frac{x^3}{2\sqrt{x^2-(1/2)^2}}\text{.} \end{equation*}

    Therefore, we use the substitution:

    \begin{equation*} x = \frac{\sec\theta}{2}, \text{ with } dx = \frac{\sec\theta\tan\theta}{2}\,d\theta\text{.} \end{equation*}

    Therefore the integral becomes:

    \begin{equation*} \begin{split} \int \frac{x^3}{2\sqrt{x^2-1/4}}\,dx \amp = \int \frac{\sec^3\theta/8}{2\sqrt{(\sec^2\theta-1)/4}} \frac{\sec\theta\tan\theta}{2}\,d\theta\\ \amp = \int \frac{\sec^3\theta/8}{\tan\theta} \frac{\sec\theta\tan\theta}{2}\,d\theta \\ \amp = \frac{1}{16} \int \sec^4\theta \,d\theta \\ \amp = \frac{1}{16(3)} \bigl(\cos(2\theta) + 2\bigr) \tan\theta \sec^2 \theta + C \\ \end{split} \end{equation*}

    We now write the integral back in terms of \(x\text{:}\) since

    we have

    \begin{equation*} \begin{split} \sec^2\theta \amp = 4x^2\\ \tan\theta \amp= \sqrt{4x^2-1}\\ \cos(2\theta) \amp = \frac{1-\sqrt{4x^2-1}}{4x^2} \end{split} \end{equation*}

    All together, we find that

    \begin{equation*} \ds\int{x^3\over\sqrt{4x^2-1}}\,dx = \frac{1}{24} (2x^2+1)\sqrt{4x^2-1}+C \end{equation*}