부분적분법

(a) We first write $$\int e^{2x} \cos(e^x) \, dx = \int e^x \cdot e^x \cos(e^x) \, dx\,.$$ Letting $z = e^x$, we have $dz = e^x \, dx$, and thus $$\int e^{2x} \cos(e^x) \, dx = \int z \cos(z) \, dz\,.$$ (b) We can evaluate the integral $\int z \cos(z) \, dz$ using integration by parts in the standard way with $u = z$ and $dv = \cos(z) \, dz$. Doing so, $$\int z \cos(z) \, dz = z \sin(z) - \int \sin(z) \, dz = z \sin(z) + \cos(z) + C\,.$$ Returning to the original integral in $x$ and using the substitution we first used with $z$, we have shown that $$\int e^{2x} \cos(e^x) \, dx = e^x \sin(e^x) + \cos(e^x) + C\,.$$ (c) For the integral $\int e^{2x} \cos(e^{2x}) \, dx$, we actually have a function-derivative pair (up to a constant multiple) already present: letting $u = e^{2x}$, $du = 2e^{2x} \, dx$, and thus we can evaluate the integral using this substitution. In particular, $$\begin{aligned}\int e^{2x} \cos(e^{2x}) \, dx &= \int \cos(u) \cdot \frac{1}{2} \, du \\ &= \frac{1}{2} \sin(u) + C \\ &= \frac{1}{2} \sin(e^{2x}) + C\end{aligned}$$ (d) - For the first integral, $\int e^{2x} \sin(e^x) \, dx$, since $e^x$ is the inside function in the composition, we let $z = e^x$, so $dz = e^x \, dx$, and write $e^{2x} = e^x \cdot e^x$. Hence, $$\int e^{2x} \sin(e^x) \, dx = \int e^x \cdot e^x \sin(e^x) \, dx = \int z \sin(z) \, dz\,.$$ Using the standard integration by parts with $u = z$ and $dv = \sin(z) \, dz$, we find $$\int z \sin(z) \, dz = \sin(z) - z\cos(z) + C\,.$$ Re-substituting from $z$ back to $x$, we've found that $$\int e^{2x} \sin(e^x) \, dx = \sin(e^x) - z\cos(e^x) + C\,.$$

• The integral $\int e^{3x} \sin(e^{3x}) \, dx$ can be evaluated using a $u$-substitution since $u = e^{3x}$ is present to form a function-derivative pair with $du = 3e^{3x}$. It follows that $$\int e^{3x} \sin(e^{3x}) \, dx = \int \sin(u) \cdot \frac{1}{3} \, du = -\frac{1}{3} \cos(u) + C = -\frac{1}{3} \cos(e^{3x}) + C\,.$$
• For the final integral, $\int xe^{x^2} \cos(e^{x^2}) \sin(e^{x^2}) \, dx$, we first observe by the chain rule that $\frac{d}{dx}\left[e^{x^2}\right] = 2xe^{x^2}$, which shows that we have a function-derivative pair present. Using the substitution $z = e^{x^2}$, it follows $dz = 2xe^{x^2} \, dx$, and therefore $$\int xe^{x^2} \cos(e^{x^2}) \sin(e^{x^2}) \, dx = \int \cos(z) \sin(z) \cdot \frac{1}{2} \, dz\,.$$ The new integral in $z$ can be evaluated using several different approaches, including $u$-substitution with $u = \sin(z)$, $du = \cos(z) \, dz$. Doing so, $$\int \cos(z) \sin(z) \cdot \frac{1}{2} \, dz = \frac{1}{2} \int u \, du = \frac{1}{2} \cdot \frac{1}{2} u^2 + C = \frac{1}{4} \sin^2(z) + C\,.$$ Returning to the original integral in $x$ and our earlier work, substituting backward we find that $$\int xe^{x^2} \cos(e^{x^2}) \sin(e^{x^2}) \, dx = \frac{1}{4} \sin^2(e^{x^2}) + C\,.$$

(a) We will use a $u$-substitution with $u = x^3$ because the integrand involves a composite function with $x^3$ as the innermost function and because $du = 3x^2 \, dx$ involves the factor $x^2$ which is also present in the integrand. When we make this substitution we will have $x^2 \ dx = \frac{1}{3} \ du$ and the integral can be evaluated as follows: $$\begin{aligned}\int x^2 \cos(x^3) \ dx &= \frac{1}{3} \int \cos(u) \ du \\ &= \frac{1}{3} \sin(u) + C \\ &= \frac{1}{3} \sin(x^3) + C\end{aligned}\,.$$ (b) We might try substituting $z = x^3/$, since there is a composite function in the integrand. In this case we have $dz = 3x^2 \, dx$. To make this substitution we need to rewrite the integrand as $$x^5 \cos(x^3) = x^3 \left[x^2 \cos(x^3) \right]$$ and notice that $x^3 = z$. Then $$\int x^5 \cos(x^3) \ dx = \frac{1}{3} \int z \cos(z) \ dz\,.$$ We can then try integration by parts on this last integral with $$\begin{aligned}u &= z, \ du = dz \\ dv & = \cos(z) \, dz, \ v = \sin(z)\end{aligned}$$ We then have $$\begin{aligned}\int x^5 \cos(x^3) \ dx &= \frac{1}{3} \int z \cos(z) \ dz \\ &= \frac{1}{3} \left(z\sin(z) - \int \sin(z) \ dz \right) \\ &= \frac{1}{3} \left(z\sin(z) + \cos(z) \right) + C \\ &= \frac{1}{3} \left(x^3\sin(x^3) + \cos(x^3) \right) + C\end{aligned}\,.$$ (c) Since the logarithm function changes to a rational function when we differentiate it, using integration by parts with $u = \ln(x^2) = 2\ln(x)$ will make the resulting integral easier to evaluate. So we let $$\begin{aligned}u &= \ln(x^2), \ du = \frac{2}{x} \ dx \\ dv & = x \ dx , \ v = \frac{x^2}{2}\end{aligned}$$ and then integration by parts yields $$\begin{aligned}\int x ln(x^2) \ dx &= \frac{x^2}{2} \ln(x^2) - \int \left(\frac{2}{x}\right) \left(\frac{x^2}{2}\right) \ dx \\ &= \frac{x^2}{2} \ln(x^2) - \int x \ dx \\ &= \frac{x^2}{2} \ln(x^2) - \frac{x^2}{2} + C\end{aligned}\,.$$ (d) We might try a substitution with $u = x^4$, but then $du = 4x^3 \ dx$ and there is no factor of $x^3$ in the integrand to allow us to complete this substitution. If we attempt to integrate by parts, then only option is to let $u = sin(x^4)$ which makes $du = 4x^3 \cos(x^4)$ and we only make the integral more complicated. This is one integral we cannot evaluate with the tools we have. (e) We will try a $u$-substitution with $u = x^4$, because the integrand involves a composite function with $x^4$ as the innermost function and because $du = 4x^3 \, dx$ involves the factor $x^3$ which is also present in the integrand. When we make this substitution we have $x^3 \ dx = \frac{1}{4} \ du$ and the integral can be evaluated as follows: $$\begin{aligned}\int x^3 \sin(x^4) \ dx &= \frac{1}{4} \int \sin(u) \ du \\ &= -\frac{1}{4} \cos(u) + C \\ &= -\frac{1}{4} \cos(x^4) + C\end{aligned}\,.$$ (f) We will use integration by parts. Notice that $$\int x^7 \sin(x^4) \ dx = \int x^4 \left[x^3 \sin(x^4)\right] \ dx$$ and we have a factor we can differentiate ($x^3$) and one we can integrate ($x^3 \sin(x^4)$) (as we did in part (e)). So we let $$\begin{aligned}u &= x^3, \ du = 3x^2 \ dx \\ dv & = x^3 \sin(x^4) \ dx , \ v = -\frac{1}{4} \cos(x^4)\end{aligned}$$ Integration by parts formula now gives $$\int x^7 \sin(x^4) \ dx = -\frac{1}{4} x^4 \cos(x^4) + \int x^3 \cos(x^4) \ dx\,.$$ This last integral is essentially the same as obtaining $v$ from $dv$, so we will make the substitution $z = x^4$ with $dz = 4x^3 \ dx$. Changing variables in the last integral gives us $$\begin{aligned}\int x^7 \sin(x^4) \ dx &= -\frac{1}{4} x^4 \cos(x^4) + \int x^3 \cos(x^4) \ dx \\ &= -\frac{1}{4} x^4 \cos(x^4) + \frac{1}{4} \int \cos(z) \ dz \\ &= -\frac{1}{4} x^4 \cos(x^4) + \frac{1}{4} \sin(z) + C \\ &= -\frac{1}{4} x^4 \cos(x^4) + \frac{1}{4} \sin(x^4) + C\end{aligned}\,.$$