We begin by noting that \(9\left(\sin^2(\theta) + \cos^2(\theta)\right) = 9\text{,}\) and hence \(9\cos^2(\theta) = 9-9\sin^2(\theta)\text{.}\) If we let \(x=3\sin(\theta)\text{,}\) then \(9-x^2 = 9-9\sin^2(\theta) = 9\cos^2(\theta)\text{.}\)

Setting \(x=3\sin(\theta)\) gives \(dx = 3\cos(\theta) \ d\theta\text{.}\) We are almost ready to substitute. We also wish to change our bounds of integration. The bound \(x=-3\) corresponds to \(\theta = -\pi/2\) (for when \(\theta = -\pi/2\text{,}\) \(x=3\sin(\theta) = -3\)). Likewise, the bound of \(x=3\) is replaced by the bound \(\theta = \pi/2\text{.}\) Thus \begin{align*} \int_{-3}^3\sqrt{9-x^2}\ dx \amp = \int_{-\pi/2}^{\pi/2} \sqrt{9-9\sin^2(\theta) }\ (3\cos(\theta) )\ d\theta\\ \amp = \int_{-\pi/2}^{\pi/2} 3\sqrt{9\cos^2(\theta) } \cos(\theta) \ d\theta\\ \amp =\int_{-\pi/2}^{\pi/2} 3\abs{3\cos(\theta) } \cos(\theta) \ d\theta.\\ \end{align*} On \([-\pi/2,\pi/2]\text{,}\) \(\cos(\theta)\) is always positive, so we can drop the absolute value bars, then employ a power--reducing formula: \begin{align*} \amp = \int_{-\pi/2}^{\pi/2} 9\cos^2(\theta) \ d\theta\\ \amp = \int_{-\pi/2}^{\pi/2} \frac{9}{2}\big(1+\cos(2\theta)\big)\ d\theta\\ \amp = \frac92 \big(\theta +\frac12\sin(2\theta)\big)\Bigg|_{-\pi/2}^{\pi/2}\\ \amp = \frac92\pi\text{.} \end{align*}

This matches our answer from before.

Using Item 3 from Key Idea 6.4.2, we recognize \(a=2\) and set \(x= 2\sec(\theta)\) for \(0 \leq \theta \lt \pi/2\text{.}\) This makes \(dx = 2\sec(\theta)\tan(\theta) \ d\theta\text{.}\) We will use the Pythagorean identity \(\tan^2(\theta)=\sec^2(\theta)-1\) and the fact that \(\tan(\theta)\geq 0\) on the interval. Substituting, we have: \begin{align*} \int \frac{1}{\sqrt{x^2-4}}\ dx \amp = \int \frac{1}{\sqrt{4\sec^2(\theta)-4}}\cdot 2 \sec(\theta)\tan(\theta)\ d\theta\\ \amp = 2\int \frac{1}{\sqrt{4(\sec^2(\theta)-1)}}\cdot \sec(\theta)\tan(\theta)\ d\theta\\ \amp = 2\int \frac{1}{\sqrt{4(\tan^2(\theta)}}\cdot\sec(\theta)\tan(\theta)\ d\theta\\ \amp = 2\int \frac{1}{2\abs{\tan(\theta)}}\sec(\theta)\tan(\theta)\ d\theta\\ \amp = \int \sec(\theta)\ d\theta\\ \amp =\ln \abs{\sec(\theta)+\tan(\theta)} \end{align*} However, our original integral involved \(x\text{.}\)

Using Item 2Key Idea 6.4.2, we recognize \(a=\sqrt{5}\) and set \(x= \sqrt{5}\tan(\theta)\text{.}\) This makes \(dx = \sqrt{5}\sec^2(\theta) \ d\theta\text{.}\) We will use the fact that \(\sqrt{5+x^2} = \sqrt{5+5\tan^2(\theta) } = \sqrt{5\sec^2(\theta) } = \sqrt{5}\sec(\theta)\text{.}\) Substituting, we have: \begin{align*} \int \frac{1}{\sqrt{5+x^2}}\ dx \amp = \int \frac{1}{\sqrt{5+5\tan^2(\theta) }}\sqrt{5}\sec^2(\theta) \ d\theta\\ \amp = \int \frac{\sqrt{5}\sec^2(\theta) }{\sqrt{5}\sec(\theta) } \ d\theta\\ \amp = \int \sec(\theta) \ d\theta\\ \amp = \ln\abs{\sec(\theta) +\tan(\theta) }+C. \end{align*}

While the integration steps are over, we are not yet done. The original problem was stated in terms of \(x\text{,}\) whereas our answer is given in terms of \(\theta\text{.}\) We must convert back to \(x\text{.}\)

The reference triangle given in Key Idea 6.4.2(b) helps. With \(x=\sqrt{5}\tan(\theta)\text{,}\) we have \begin{equation*} \tan(\theta) = \frac x{\sqrt{5}} \text{ and } \sec(\theta) = \frac{\sqrt{x^2+5}}{\sqrt{5}}. \end{equation*}

This gives \begin{align*} \int \frac{1}{\sqrt{5+x^2}}\ dx \amp = \ln\abs{\sec(\theta) +\tan(\theta) }+C\\ \amp = \ln\abs{\frac{\sqrt{x^2+5}}{\sqrt{5}}+ \frac x{\sqrt{5}}}+C. \end{align*}

We can leave this answer as is, or we can use a logarithmic identity to simplify it. Note: \begin{align*} \ln\abs{\frac{\sqrt{x^2+5}}{\sqrt{5}}+ \frac x{\sqrt{5}}}+C \amp = \ln\abs{\frac{1}{\sqrt{5}}\big(\sqrt{x^2+5}+ x\big)}+C\\ \amp = \ln\abs{\frac{1}{\sqrt{5}}} + \ln\abs{\sqrt{x^2+5}+ x}+C\\ \amp = \ln\abs{\sqrt{x^2+5}+ x}+C, \end{align*} where the \(\ln\big(1/\sqrt{5}\big)\) term is absorbed into the constant \(C\text{.}\) (In Section 6.6 we will learn another way of approaching this problem.)

We start by rewriting the integrand so that it looks like \(\sqrt{x^2-a^2}\) for some value of \(a\text{:}\) \begin{align*} \sqrt{4x^2-1} \amp = \sqrt{4\left(x^2-\frac14\right)}\\ \amp = 2\sqrt{x^2-\left(\frac12\right)^2}. \end{align*}

So we have \(a=1/2\text{,}\) and following Key Idea 6.4.2(c), we set \(x= \frac12\sec(\theta)\text{,}\) and hence \(dx = \frac12\sec(\theta) \tan(\theta) \ d\theta\text{.}\) We now rewrite the integral with these substitutions: \begin{align*} \int \sqrt{4x^2-1}\ dx \amp = \int 2\sqrt{x^2-\left(\frac12\right)^2}\ dx\\ \amp = \int 2\sqrt{\frac14\sec^2(\theta) - \frac14}\left(\frac12\sec(\theta) \tan(\theta) \right)\ d\theta\\ \amp =\int \sqrt{\frac14(\sec^2(\theta) -1)}\Big(\sec(\theta) \tan(\theta) \Big)\ d\theta\\ \amp =\int\sqrt{\frac14\tan^2(\theta) }\Big(\sec(\theta) \tan(\theta) \Big)\ d\theta\\ \amp =\int \frac12\tan^2(\theta) \sec(\theta) \ d\theta\\ \amp =\frac12\int \Big(\sec^2(\theta) -1\Big)\sec(\theta) \ d\theta\\ \amp =\frac12\int \big(\sec^3(\theta) - \sec(\theta) \big)\ d\theta. \end{align*}

We integrated \(\sec^3(\theta)\) in Example 6.3.10, finding its antiderivatives to be \begin{equation*} \int \sec^3(\theta) \ d\theta = \frac12\Big(\sec(\theta) \tan(\theta) + \ln\abs{\sec(\theta) +\tan(\theta) }\Big)+C. \end{equation*}

Thus \begin{align*} \int \sqrt{4x^2-1}\ dx \amp =\frac12\int \big(\sec^3(\theta) - \sec(\theta) \big)\ d\theta\\ \amp = \frac12\left(\frac12\Big(\sec(\theta) \tan(\theta) + \ln\abs{\sec(\theta) +\tan(\theta) }\Big) -\ln\abs{\sec(\theta) + \tan(\theta) }\right) + C\\ \amp = \frac14\left(\sec(\theta) \tan(\theta) -\ln\abs{\sec(\theta) +\tan(\theta) }\right)+C. \end{align*}

We are not yet done. Our original integral is given in terms of \(x\text{,}\) whereas our final answer, as given, is in terms of \(\theta\text{.}\) We need to rewrite our answer in terms of \(x\text{.}\) With \(a=1/2\text{,}\) and \(x=\frac12\sec(\theta)\text{,}\) the reference triangle in Key Idea 6.4.2(c) shows that \begin{equation*} \tan(\theta) = \sqrt{x^2-1/4}\Big/(1/2) = 2\sqrt{x^2-1/4} \text{ and } \sec(\theta) = 2x. \end{equation*}

Thus \begin{align*} \frac14\Big(\sec(\theta) \tan(\theta) -\ln\abs{\sec(\theta) +\tan(\theta) }\Big)+C \amp = \frac14\Big(2x\cdot 2\sqrt{x^2-1/4} - \ln\abs{2x + 2\sqrt{x^2-1/4}}\Big)+C\\ \amp = \frac14\Big(4x\sqrt{x^2-1/4} - \ln\abs{2x + 2\sqrt{x^2-1/4}}\Big)+C. \end{align*}

The final answer is given in the last line above, repeated here: \begin{equation*} \int \sqrt{4x^2-1}\ dx = \frac14\Big(4x\sqrt{x^2-1/4} - \ln\abs{2x + 2\sqrt{x^2-1/4}}\Big)+C. \end{equation*}

We use Key Idea 6.4.2(a) with \(a=2\text{,}\) \(x=2\sin(\theta)\text{,}\) \(dx = 2\cos(\theta)\) and hence \(\sqrt{4-x^2} = 2\cos(\theta)\text{.}\) This gives \begin{align*} \int \frac{\sqrt{4-x^2}}{x^2}\ dx \amp = \int \frac{2\cos(\theta) }{4\sin^2(\theta) }(2\cos(\theta) )\ d\theta\\ \amp = \int \cot^2(\theta) \ d\theta\\ \amp = \int (\csc^2(\theta) -1)\ d\theta\\ \amp = -\cot(\theta) -\theta + C. \end{align*}

We need to rewrite our answer in terms of \(x\text{.}\) Using the reference triangle found in Key Idea 6.4.2(a), we have \(\cot(\theta) = \sqrt{4-x^2}/x\) and \(\theta = \sin^{-1}(x/2)\text{.}\) Thus \begin{equation*} \int \frac{\sqrt{4-x^2}}{x^2}\ dx = -\frac{\sqrt{4-x^2}}x-\sin^{-1}\left(\frac x2\right) + C. \end{equation*}

We know the answer already as \(\tan^{-1}(x) +C\text{.}\) We apply Trigonometric Substitution here to show that we get the same answer without inherently relying on knowledge of the derivative of the arctangent function.

Using Key Idea 6.4.2(b), let \(x=\tan(\theta)\text{,}\) \(dx=\sec^2(\theta) \ d\theta\) and note that \(x^2+1 = \tan^2(\theta) +1 = \sec^2(\theta)\text{.}\) Thus \begin{align*} \int \frac1{x^2+1}\ dx \amp = \int \frac{1}{\sec^2(\theta) }\sec^2(\theta) \ d\theta\\ \amp = \int 1\ d\theta\\ \amp = \theta + C. \end{align*}

Since \(x=\tan(\theta)\text{,}\) \(\theta = \tan^{-1}(x)\text{,}\) and we conclude that \(\ds \int\frac1{x^2+1}\ dx = \tan^{-1}(x) +C\text{.}\)

We start by completing the square, then make the substitution \(u=x+3\text{,}\) followed by the trigonometric substitution of \(u=\tan(\theta)\text{:}\) \begin{align*} \int \frac1{(x^2+6x+10)^2}\ dx =\int \frac1{\big((x+3)^2+1\big)^2}\ dx\amp = \int \frac1{(u^2+1)^2}\ du.\\ \end{align*} Now make the substitution \(u=\tan(\theta)\text{,}\) \(du=\sec^2(\theta) \ d\theta\text{:}\) \begin{align*} \amp = \int \frac1{(\tan^2(\theta) +1)^2}\sec^2(\theta) \ d\theta\\ \amp = \int\frac 1{(\sec^2(\theta) )^2}\sec^2(\theta) \ d\theta\\ \amp = \int \cos^2(\theta) \ d\theta.\\ \end{align*} Applying a power reducing formula, we have \begin{align*} \amp = \int \left(\frac12 +\frac12\cos(2\theta)\right)\ d\theta\\ \amp = \frac12\theta + \frac14\sin(2\theta) + C. \end{align*}

We need to return to the variable \(x\text{.}\) As \(u=\tan(\theta)\text{,}\) \(\theta = \tan^{-1}(u)\text{.}\) Using the identity \(\sin(2\theta) = 2\sin(\theta) \cos(\theta)\) and using the reference triangle found in Key Idea 6.4.2(b), we have \begin{equation*} \frac14\sin(2\theta) = \frac12\frac u{\sqrt{u^2+1}}\cdot\frac 1{\sqrt{u^2+1}} = \frac12\frac u{u^2+1}. \end{equation*}

Finally, we return to \(x\) with the substitution \(u=x+3\text{.}\) We start with the expression in Equation <<Unresolved xref, reference "eq_extrigsub7"; check spelling or use "provisional" attribute>>: \begin{align*} \frac12\theta + \frac14\sin(2\theta) + C \amp = \frac12\tan^{-1}(u) + \frac12\frac{u}{u^2+1}+C\\ \amp = \frac12\tan^{-1}(x+3) + \frac{x+3}{2(x^2+6x+10)}+C. \end{align*}

Stating our final result in one line, \begin{equation*} \int\frac1{(x^2+6x+10)^2}\ dx=\frac12\tan^{-1}(x+3) + \frac{x+3}{2(x^2+6x+10)}+C. \end{equation*}

Using Key Idea 6.4.2(b), we set \(x=5\tan(\theta)\text{,}\) \(dx = 5\sec^2(\theta) \ d\theta\text{,}\) and note that \(\sqrt{x^2+25} = 5\sec(\theta)\text{.}\) As we substitute, we can also change the bounds of integration.

The lower bound of the original integral is \(x=0\text{.}\) As \(x=5\tan(\theta)\text{,}\) we solve for \(\theta\) and find \(\theta = \tan^{-1}(x/5)\text{.}\) Thus the new lower bound is \(\theta = \tan^{-1}(0) = 0\text{.}\) The original upper bound is \(x=5\text{,}\) thus the new upper bound is \(\theta = \tan^{-1}(5/5) = \pi/4\text{.}\)

Thus we have \begin{align*} \int_0^5\frac{x^2}{\sqrt{x^2+25}}\ dx \amp = \int_0^{\pi/4} \frac{25\tan^2(\theta) }{5\sec(\theta) }5\sec^2(\theta) \ d\theta\\ \amp = 25\int_0^{\pi/4} \tan^2(\theta) \sec(\theta) \ d\theta. \end{align*}

We encountered this indefinite integral in Example 6.4.5 where we found \begin{equation*} \int \tan^2(\theta) \sec(\theta) \ d\theta = \frac12\big(\sec(\theta) \tan(\theta) -\ln\abs{\sec(\theta) +\tan(\theta) }\big). \end{equation*}

So \begin{align*} 25\int_0^{\pi/4} \tan^2(\theta) \sec(\theta) \ d\theta \amp = \frac{25}2\big(\sec(\theta) \tan(\theta) -\ln\abs{\sec(\theta) +\tan(\theta) }\big)\Bigg|_0^{\pi/4}\\ \amp = \frac{25}2\big(\sqrt2-\ln(\sqrt2+1)\big)\\ \amp \approx 6.661. \end{align*}