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