Let computing \(du\), we find and solving for \(dx\) we find Now \begin{align*} \int _{2}^{3} \frac {1}{x\ln (x)} dx &= \int _{g(2)}^{g(3)} \frac {1}{x \cdot u} x du \\ &= \int _{\answer [given]{\ln (2)}}^{\answer [given]{\ln (3)}} \answer [given]{\frac {1}{u}} du\\ &= \bigg [ \answer [given]{\ln (u)} \bigg ]_{\answer [given]{\ln (2)}}^{\answer [given]{\ln (3)}}\\ & = \ln (\ln (3)) - \ln (\ln (2)). \end{align*}

Here it is not apparent that the chain rule is involved. However, if it was involved, perhaps a good guess for \(g\) would be and then \begin{align*} du &= \answer [given]{-2x} dx, \\ dx &= \answer [given]{\frac {-1}{2x}} du. \end{align*}

Now we consider the integral we are trying to compute

and we substitute using our work above. Write with me \begin{align*} \int x^3\sqrt {1-x^2} dx &= \int x^3 \answer [given]{\sqrt {u}} \left ( \answer [given]{\frac {-1}{2x}} \right ) du \\ &= \int \frac {-x^2 \answer [given]{\sqrt {u}}}{2} du. \end{align*}

However, we cannot continue until each \(x\) is replaced. We know that \begin{align*} u &= 1-x^2 \\ \Rightarrow \qquad u -1 &= -x^2\\ \Rightarrow \qquad \answer [given]{1- u} &= x^2 \end{align*}

so now we may replace \(x^2\)

At this point, we are close to being done. Write \begin{align*} \int -\frac {(1-u)\sqrt {u}}{2} du &= \int \left (\frac {u\sqrt {u}}{2} - \frac {\sqrt {u}}{2}\right ) du \\ &= \int \frac {\answer [given]{u^{3/2}}}{2} du - \int \frac {\sqrt {u}}{2} du \\ &= \answer [given]{\frac {u^{5/2}}{5}} - \answer [given]{\frac {u^{3/2}}{3}}. \end{align*}

Now recall that \(u = 1-x^2\). Hence our final answer is

We substitute and we immediately see that \begin{align*} du &= \answer [given]{(\sec (y) \tan (y) + \sec ^2(y))} dy,\\ dy &= \frac {1}{\answer [given]{\sec (y) \tan (y) + \sec ^2(y)}} du. \end{align*}

But this cancels perfectly with the numerator! So we have that \begin{align*} \int \frac {\sec (y) \tan (y) + \sec ^2(y)}{\sec (y) + \tan (y)} dy &= \int \frac {1}{u} du \\ &= \ln |u| + C \\ &= \ln | \sec (y) + \tan (y) | + C. \end{align*}

We want to substitute for \(1-u^2\). But the variable “\(u\)” has already been used…OH NO! Never fear! We can substitute with whatever variable that we want. In particular, let us use “\(w\)” for this problem. So we let and then \begin{align*} dw &= \answer [given]{-2u} du,\\ du &= \answer [given]{\frac {-1}{2u}} dw. \end{align*}

Thus \begin{align*} \int \frac {u}{1-u^2} du &= \int \frac {u}{w} \left ( \answer [given]{\frac {-1}{2u}} \right ) dw \\ &= - \frac {1}{2} \int \frac {1}{w} dw \\ &= - \frac {1}{2} \ln |w| + C \\ &= - \frac {1}{2} \ln |1-u^2| + C. \end{align*}

We begin by writing \begin{align*} \tan (x) &= \frac {\sin (x)}{\cos (x)} \\ &= \frac {\sin (x)}{\cos (x)} \cdot \frac {\cos (x)}{\cos (x)} \\ &= \frac {\sin (x) \cos (x)}{1 - \sin ^2(x)}. \end{align*}

We then make the substitution

and so \begin{align*} ds &= \answer [given]{\cos (x)} dx,\\ dx &= \frac {1}{\answer [given]{\cos (x)}} ds. \end{align*}

Then \begin{align*} \int \tan (x) dx &= \int \frac {\sin (x) \cos (x)}{1 - \sin ^2(x)} dx \\ &= \int \frac {s \cos (x)}{1 - s^2} \cdot \frac {1}{\answer [given]{\cos (x)}} ds \\ &= \int \frac {s}{1-s^2} ds. \end{align*}

But this is the same problem as Example key example! And so we know that \begin{align*} \int \tan (x) dx &= - \frac {1}{2} \ln |1-s^2| + C \\ &= - \frac {1}{2} \ln |1-\sin ^2 (x)| + C \\ &= - \frac {1}{2} \ln |\cos ^2(x)| + C \\ &= \ln |\cos ^2(x)|^{- \frac {1}{2}} + C \\ &= \ln |\sec (x)| + C. \end{align*}

Maybe the biggest key to solving this problem is to recall that So we can rewrite the problem Now, if we make the substitution \(u = e^x\), we have that \begin{align*} du &= \answer [given]{e^x} dx,\\ dx &= \frac {1}{\answer [given]{e^x}} du, \end{align*}

and \begin{align*} \int \frac {e^{2x}}{1 - e^{2x}} dx &= \int \frac {(e^x)^2}{1 - u^2} \cdot \frac {1}{\answer [given]{e^x}} du \\ &= \int \frac {e^x}{1-u^2} du \\ &= \int \frac {u}{1-u^2} du. \end{align*}

But now we are back to Example key example, and so we know that \begin{align*} \int \frac {e^{2x}}{1 - e^{2x}} dx &= - \frac {1}{2} \ln |1-u^2| + C \\ &= - \frac {1}{2} \ln |1 - e^{2x}| + C. \end{align*}

While it is not obvious at all, let us try the substitution Then \begin{align*} du &= \answer [given]{\frac {1}{2 \sqrt {x}}} dx,\\ dx &= 2 \sqrt {x} du = 2u du, \end{align*}

and so \begin{align*} \int _0^{16} \sqrt {4 - \sqrt {x}} dx &= \int _{g(0)}^{g(16)} \answer [given]{\sqrt {4-u} \cdot 2u} du \\ &= \int _{\answer [given]{0}}^{\answer [given]{4}} \answer [given]{2u \sqrt {4-u}} du. \end{align*}

From here we now make the second (and more obvious) substitution

Then \(u = 4-w\), and \begin{align*} dw &= - du,\\ du &= - dw. \end{align*}

So \begin{align*} \int _0^{16} \sqrt {4 - \sqrt {x}} dx &= \int _0^4 2u \sqrt {4-u} du \\ &= \int _{w(0)}^{w(4)} 2 (4-w) \sqrt {w} (-1) dw \\ &= - \int _{\answer [given]{4}}^{\answer [given]{0}} ( 8w^{\frac {1}{2}} - 2w^{\frac {3}{2}} ) dw \\ &= \int _{\answer [given]{0}}^{\answer [given]{4}} ( 8w^{\frac {1}{2}} - 2w^{\frac {3}{2}} ) dw \\ &= \bigg [ \answer [given]{8 \left ( \frac {2}{3} \right ) w^{\frac {3}{2}} - 2 \left ( \frac {2}{5} \right ) w^{\frac {5}{2}}} \bigg ]_{0}^{4} \\ &= \left ( \frac {16}{3} (4)^{\frac {3}{2}} - \frac {4}{5} (4)^{\frac {5}{2}} \right ) - \left ( 0 - 0 \right ) \\ &= \frac {128}{3} - \frac {128}{5} \\ &= \frac {256}{15}. \end{align*}