Coordinated Calculus

When we use integration by parts, we have a choice for \(u\) and \(dv\text{.}\) In this problem, we can either let \(u = x\) and \(dv = \cos(x) \, dx\text{,}\) or let \(u = \cos(x)\) and \(dv = x \, dx\text{.}\) While there is not a universal rule for how to choose \(u\) and \(dv\text{,}\) a good guideline is this: do so in a way that \(\int v \, du\) is at least as simple as the original problem \(\int u \, dv\text{.}\)

This leads us to choose⁷Observe that if we considered the alternate choice, and let \(u = \cos(x)\) and \(dv = x \, dx\text{,}\) then \(du = -\sin(x) \, dx\) and \(v = \frac{1}{2}x^2\text{,}\) from which we would write \(\int x\cos(x) \, dx = \frac{1}{2}x^2 \cos(x) - \int \frac{1}{2}x^2 (-\sin(x)) \, dx\text{.}\) Thus we have replaced the problem of integrating \(x \cos(x)\) with that of integrating \(\frac{1}{2}x^2 \sin(x)\text{;}\) the latter is clearly more complicated, which shows that this alternate choice is not as helpful as the first choice.\(u = x\) and \(dv = \cos(x) \, dx\text{,}\) from which it follows that \(du = 1 \, dx\) and \(v = \sin(x)\text{.}\) With this substitution, the rule for integration by parts tells us that

All that remains to do is evaluate the (simpler) integral \(\int \sin(x) \cdot 1 \, dx\text{.}\) Doing so, we find

Observe that when we get to the final stage of evaluating the last remaining antiderivative, it is at this step that we include the integration constant, \(+C\text{.}\)

1. Try \(u=t\text{.}\)

2. Let \(dv=\sin(3x)dx\)

3. Remember that \(\int \sec^2(z) \,dz = \tan(z)\text{.}\)

4. Note that \(\ln(x) \, dx\) has a simple derivative to work with.

1. \(\int t e^{-t} dt = -te^{-t} - e^{-t} +C\text{.}\)

2. \(\int 4x \sin(3x) dx = -\dfrac{4}{3} x \cos(3x) + \frac{4}{9} \sin(3x) + C \text{.}\)

3. \(\int z \sec^2(z) dz = z \tan(z) + \ln |\cos(z)| + C \text{.}\)

4. \(\int x\ln(x) dx = \frac{1}{2}x^2 \ln(x) - \frac{1}{4}x^2 + C \text{.}\)

1. Using \(u = t\) and \(dv = e^{-t} dt\text{,}\) we obtain \(du = dt\) and \(v = -e^{-t}\text{.}\) So

   \begin{align*} \int t e^{-t} dt \amp = -te^{-t} + \int e^{-t} dt\\ \amp = -te^{-t} - e^{-t} + C \end{align*}

   We check using differentiation as follows:

   \begin{align*} \frac{d}{dt} \left( -te^{-t} - e^{-t} + C \right) \amp = \left( t e^{-t} - e^{-t} \right) + e^{-t}\\ \amp = te^{-t} \end{align*}

2. Using \(u = 4x\) and \(dv = \sin(3x) dx\text{,}\) we obtain \(du = 4 dx\) and \(v = -\dfrac{1}{3} \cos(3x)\text{.}\) So

   \begin{align*} \int 4x \sin(3x) dx \amp = -\dfrac{4}{3} x \cos(3x) + \dfrac{4}{3} \int \cos(3x) dx\\ \amp = -\dfrac{4}{3} x \cos(3x) + \frac{4}{9} \sin(3x) + C \end{align*}

   We check using differentiation as follows:

   \begin{align*} \frac{d}{dx} \left( -\dfrac{4}{3} x \cos(3x) + \frac{4}{9} \sin(3x) + C \right) \amp = 4x \sin(3x) - \frac{4}{3} \cos(3x) + \frac{4}{3} \cos(3x)\\ \amp = 4x \sin(3x) \end{align*}

3. Using \(u = z\) and \(dv = \sec^2(z) dz\text{,}\) we obtain \(du = dz\) and \(v = \tan(z)\text{.}\) So

   \begin{align*} \int z \sec^2(z) dz \amp = z \tan(z) - \int \tan(z) \,dz\\ \int z \sec^2(z) dz \amp = z \tan(z) - \int \frac{\sin(z)}{\cos(z)} \,dz\\ \amp = z \tan(z) + \ln |\cos(z)| + C \end{align*}

   We check using differentiation as follows:

   \begin{align*} \frac{d}{dz} \left( z \tan(z) + \ln |\cos(z)| + C \right) \amp = z\sec^2(z) + \tan(z) - \frac{\sin(z)}{\cos(z)}\\ \amp = z\sec^2(z) + \tan(z) - \tan(z)\\ \amp = z \sec^2(z) \end{align*}

4. Using \(u = \ln(x)\) and \(dv = x dx\text{,}\) we obtain \(du = \dfrac{1}{x} dx\) and \(v = \dfrac{1}{2}x^2\text{.}\) So

   \begin{align*} \int x\ln(x) dx \amp = \frac{1}{2}x^2 \ln(x) - \frac{1}{2} \int x dx\\ \amp = \frac{1}{2}x^2 \ln(x) - \frac{1}{4}x^2 + C \end{align*}

   We check using differentiation as follows:

   \begin{align*} \frac{d}{dx} \left( \frac{1}{2}x^2 \ln(x) - \frac{1}{4}x^2 + C \right) \amp = \frac{1}{2}x + x \ln(x) - \frac{1}{2} x\\ \amp = x \ln(x) \end{align*}

1. \(\int{\arctan(x) dx} = x\arctan(x) - \frac{1}{2} \ln \left( 1 + x^2 \right) + C \text{.}\)

2. \(\int \ln(z) dz = z \ln(z) - z + C \text{.}\)

3. \(\int t^3 \sin(t^2) dt = \frac{1}{2} \left( -t^2 \cos\left(t^2 \right) + \sin\left(t^2\right) \right) +C\text{.}\)

4. \(\int s^5 e^{s^3} ds = \frac{1}{3} \left( s^3 e^{s^3} - e^{s^3} \right) + C \text{.}\)

5. \(\int e^{2t} \cos\left( e^t \right) dt = e^t \sin \left( e^t \right) + \cos \left( e^t \right) + C \text{.}\)

1. We use integration by parts with \(u = \arctan(x)\) and \(dv = dx\text{.}\) So \(du = \dfrac{1}{1+x^2}dx\) and \(v = x\text{.}\)

   \begin{align*} \int{\arctan(x) dx} \amp = uv - \int{v du}\\ \amp = x\arctan(x) - \int\frac{x}{1+x^2} dx \end{align*}

   For the integral in the last equation, use the substitution \(z = 1+x^2\) and \(dz = 2x\, dx\text{.}\)

   \begin{align*} \int \arctan(x) dx \amp = x\arctan(x) -\int\frac{x}{1+x^2} dx\\ \amp = x\arctan(x) - \frac{1}{2} \int \frac{1}{z} dz\\ \amp = x\arctan(x) - \frac{1}{2} \ln(z) + C\\ \amp = x\arctan(x) - \frac{1}{2} \ln \left( 1 + x^2 \right) + C \end{align*}

2. Using \(u = \ln(z)\) and \(dv = dz\text{,}\) we obtain \(du = \dfrac{1}{z} dz\) and \(v = z\text{.}\) So

   \begin{align*} \int \ln(z) dz \amp = z \ln(z) - \int dz\\ \amp = z \ln(z) - z + C \end{align*}

3. Using \(z = t^2\text{,}\) we obtain \(dz = 2t dt\) and \(\dfrac{1}{2} dz = t dt\text{.}\) So

   \begin{align*} \int t^3 \sin(t^2) dt \amp = \int t^2 \sin(t^2) \cdot t dt\\ \amp = \int z \sin(z) \frac{1}{2} dz\\ \amp = \frac{1}{2} \int z \sin(z) dz \end{align*}

   Now using integration by parts with \(u = z\) and \(dv = \sin(z) dz\text{,}\) we obtain \(du = dz\) and \(v = -\cos(z)\text{.}\) So

   \begin{align*} \int t^3 \sin(t^2) dt \amp = \frac{1}{2} \int z \sin(z) dz\\ \amp = \frac{1}{2} \left( -z \cos(z) + \int \cos(z) dz \right)\\ \amp = \frac{1}{2} \left( -z \cos(z) + \sin(z) \right) + C \end{align*}

   Finally, we use \(z = t^2\) and obtain

   \begin{equation*} \int t^3 \sin(t^2) dt = \frac{1}{2} \left( -t^2 \cos\left(t^2 \right) + \sin\left(t^2\right) \right) +C\text{.} \end{equation*}

4. Use a substitution with \(z = s^3\) and \(dz = 3s^2 ds\text{.}\)

   \begin{align*} \int s^5 e^{s^3} ds \amp = \int s^2 s^3 e^{s^3} ds\\ \amp = \frac{1}{3} \int z e^z dz \end{align*}

   Now use integration by parts with \(u = z\) and \(dv = e^z dz\text{.}\) So \(du = dz\) and \(v = e^z\text{.}\)

   \begin{align*} \int s^5 e^{s^3} ds \amp = \frac{1}{3} \int z e^z dz\\ \amp = \frac{1}{3} \left( z e^z - \int e^z dz \right)\\ \amp = \frac{1}{3} \left( z e^z - e^z \right) + C\\ \amp = \frac{1}{3} \left( s^3 e^{s^3} - e^{s^3} \right) + C \end{align*}

5. Using \(z = e^t\text{,}\) we see that \(dz = e^t dt\) and so

   \begin{align*} \int e^{2t} \cos\left( e^t \right) dt \amp = \int e^t \cos\left( e^t \right) \cdot e^t dt\\ \amp = \int z \cos(z) dz \end{align*}

   In Example5.47, we saw that

   \begin{equation*} \int z \cos(z) dz = z \sin(z) + \cos(z) + C\text{.} \end{equation*}

   Combining these two results, we obtain

   \begin{align*} \int e^{2t} \cos\left( e^t \right) dt \amp = \int z \cos(z) dz\\ \amp = z \sin(z) + \cos(z) + C\\ \amp = e^t \sin \left( e^t \right) + \cos \left( e^t \right) + C \end{align*}

Let \(u = t^2\) and \(dv = e^t \, dt\text{.}\) Then \(du = 2t \, dt\) and \(v = e^t\text{,}\) and thus

The integral on the right side is simpler to evaluate than the one on the left, but it still requires integration by parts. Now letting \(u = 2t\) and \(dv = e^t \, dt\text{,}\) we have \(du = 2\, dt\) and \(v = e^t\text{,}\) so that

(Note the parentheses, which remind us to distribute the minus sign to the entire value of the integral \(\int 2t e^t \, dt\text{.}\)) The final integral on the right is a basic one; evaluating that integral and distributing the minus sign, we find

Apply integration by parts twice, then rearrange the equation to solve for \(\int e^t \cos(t) \, dt\text{.}\)

\(\int e^t \cos(t) \, dt= \frac{1}{2}(e^t\cos(t)+e^t\sin(t)) +C\text{.}\)

We can choose to let \(u\) be either \(e^t\) or \(\cos(t)\text{;}\) we pick \(u = \cos(t)\text{,}\) and thus \(dv = e^t \, dt\text{.}\) With \(du = -\sin(t) \, dt\) and \(v = e^t\text{,}\) integration by parts tells us that

or equivalently that

The new integral has the same algebraic structure as the original one. While the overall situation isn't necessarily better than what we started with, it hasn't gotten worse. Thus, we proceed to integrate by parts again. This time we let \(u = \sin(t)\) and \(dv = e^t \, dt\text{,}\) so that \(du = \cos(t) \, dt\) and \(v = e^t\text{,}\) which implies

We seem to be back where we started, as two applications of integration by parts has led us back to the original problem, \(\int e^t \cos(t) \, dt\text{.}\) But if we look closely at Equation(5.10), we see that we can use algebra to solve for the value of the desired integral. Adding \(\int e^t \cos(t) \, dt\) to both sides of the equation, we have

and therefore

Note that since we never actually encountered an integral we could evaluate directly, we didn't have the opportunity to add the integration constant \(C\) until the final step.

1. Start with \(u=x^2\text{.}\)

2. Begin with \(u=t^3\text{.}\)

3. You'll have to integrate by parts twice.

4. Start with \(u=s^2\text{.}\)

5. Try \(u=\arctan(x)\text{.}\) At a certain point in this problem, it is very helpful to note that \(\frac{t^2}{1+t^2} = 1 - \frac{1}{1+t^2}\text{.}\) Just like while using partial fractions if the numerator and denominator of a rational function are the same, performing polynomial long division first can help.

1. \begin{gather*} \int x^2 \sin(x) dx = -x^2 \cos(x) + 2x \sin(x) + 2 \cos(x) + C \text{.} \end{gather*}
2. \begin{gather*} \int t^3 \ln(t) dt = \frac{1}{4} t^4 \ln(t) - \frac{1}{16} t^4 + C \text{.} \end{gather*}
3. \begin{gather*} \int e^z \sin(z) dz = -\frac{1}{2}e^z \cos(z) + \frac{1}{2}e^z \sin(z) + C \text{.} \end{gather*}
4. \begin{gather*} \int s^2 e^{3s} ds = \frac{1}{3}s^2 e^{3s} - \frac{2}{9}s e^{3s} + \frac{2}{27} e^{3s} + C \text{.} \end{gather*}
5. \begin{align*} \int t \arctan(t) dt \amp= \frac{1}{2}t^2 \arctan(t) - \frac{1}{2}t - \frac{1}{2} \arctan(t) + C \text{.} \end{align*}

1. First, using \(u = x^2\) and \(dv = \sin(x) dx\text{,}\) we get \(du = 2x dx\) and \(v = -\cos(x)\text{.}\) So

   \begin{equation*} \int x^2 \sin(x) dx = -x^2 \cos(x) + 2 \int x \cos(x) dx\text{.} \end{equation*}

   Now, for \(\int x \cos(x) dx\text{,}\) we use \(u = x\) and \(dv = \cos(x)\) and get \(du = dx\) and \(v =\sin(x)\text{.}\) So

   \begin{align*} \int x^2 \sin(x) dx \amp = -x^2 \cos(x) + 2 \left( x \sin(x) - \int \sin(x) dx \right)\\ \amp = -x^2 \cos(x) + 2x \sin(x) + 2 \cos(x) + C \end{align*}

2. Using \(u = \ln(t)\) and \(dv = t^3 dt\text{,}\) we get \(du = \dfrac{1}{t} dt\) and \(v = \dfrac{1}{4} t^4\text{.}\) So

   \begin{align*} \int t^3 \ln(t) dt \amp = \frac{1}{4} t^4 \ln(t) - \frac{1}{4} \int t^3 dt\\ \amp = \frac{1}{4} t^4 \ln(t) - \frac{1}{4} \cdot \frac{1}{4} t^4 + C\\ \amp = \frac{1}{4} t^4 \ln(t) - \frac{1}{16} t^4 + C \end{align*}

3. We use \(u = e^z\) and \(dv = \sin(z) dz\text{.}\) This gives \(du = e^z dz\) and \(v = -\cos(z)\text{,}\) and

   \begin{equation*} \int e^z \sin(z) dz = -e^z \cos(z) + \int e^z \cos(z) dz\text{.} \end{equation*}

   For \(\int e^z \cos(z) dz\text{,}\) we use \(u = e^z\) and \(dv = \cos(z) dz\text{.}\) So \(du = e^z dz\) and \(v = \sin(z)\text{,}\) and

   \begin{align*} \int e^z \sin(z) dz \amp = -e^z \cos(z) + e^z \sin(z) dz - \int e^z \sin(z) dz\\ 2 \int e^z \sin(z) dz \amp = -e^z \cos(z) + e^z \sin(z) + C'\\ \int e^z \sin(z) dz \amp = -\frac{1}{2}e^z \cos(z) + \frac{1}{2}e^z \sin(z) + C \end{align*}

4. We use \(u = s^2\text{,}\) \(dv = e^{3s} ds\text{,}\) \(du = 2s ds\text{,}\) and \(v = \frac{1}{3} e^{3s}\text{.}\) This gives

   \begin{equation*} \int s^2 e^{3s} ds = \frac{1}{3}s^2 e^{3s} - \frac{2}{3} \int s e^{3s} ds\text{.} \end{equation*}

   We now use \(u = s\text{,}\) \(dv = e^{3s} ds\text{,}\) \(du = ds\text{,}\) and \(v = \dfrac{1}{3} e^{3s}\) for \(\int s e^{3s} ds\text{.}\)

   \begin{align*} \int s^2 e^{3s} ds \amp = \frac{1}{3}s^2 e^{3s} - \frac{2}{3} \left( \frac{1}{3}s e^{3s} - \frac{1}{3}\int e^{3s} ds \right)\\ \amp = \frac{1}{3}s^2 e^{3s} - \frac{2}{3} \left( \frac{1}{3}s e^{3s} - \frac{1}{9} e^{3s} \right) + C\\ \amp = \frac{1}{3}s^2 e^{3s} - \frac{2}{9}s e^{3s} + \frac{2}{27} e^{3s} + C \end{align*}

5. Using \(u = \arctan(t)\text{,}\) \(dv = t dt\text{,}\) \(du = \dfrac{1}{1+t^2} dt\text{,}\) and \(v = \dfrac{1}{2} t^2\text{,}\) we get

   \begin{align*} \int t \arctan(t) dt \amp = \frac{1}{2}t^2 \arctan(t) - \frac{1}{2} \int \frac{t^2}{1 + t^2} dt\\ \amp = \frac{1}{2}t^2 \arctan(t) - \frac{1}{2} \int \left( 1 - \frac{1}{1 + t^2} \right) dt\\ \amp = \frac{1}{2}t^2 \arctan(t) - \frac{1}{2}t - \frac{1}{2} \arctan(t) + C \end{align*}

1. Take \(du=\sin(t)dt\text{.}\) What is \(v\text{?}\)
2. Take \(v=e^{2t}\text{.}\) What is \(u\text{?}\)

1. \(1 \)
2. \(\sim 39.1\)

1. One option is to find an antiderivative (using indefinite integral notation) and then apply the Fundamental Theorem of Calculus to find that

   \begin{align*} \int_0^{\pi/2} t\sin(t) \, dt =\mathstrut \amp \left.\left( -t \cos(t) + \sin(t) \right) \right|_0^{\pi/2}\\ =\mathstrut \amp \left( -\frac{\pi}{2} \cos(\frac{\pi}{2}) + \sin(\frac{\pi}{2}) \right) - \left( -0 \cos(0) + \sin(0) \right)\\ =\mathstrut \amp 1\text{.} \end{align*}

   Alternatively, we can apply integration by parts and work with definite integrals throughout. With this method, we must remember to evaluate the product \(uv\) over the given limits of integration. Using the substitution \(u = t\) and \(dv = \sin(t) \, dt\text{,}\) so that \(du = dt\) and \(v = -\cos(t)\text{,}\) we write

   \begin{align*} \int_0^{\pi/2} t\sin(t) \, dt =\mathstrut\amp \left. -t \cos(t) \right |_{0}^{\pi/2} - \int_0^{\pi/2} (-\cos(t)) \, dt \\ =\mathstrut \amp \left. -t \cos(t) \right |_{0}^{\pi/2} + \left. \sin(t) \right |_{0}^{\pi/2} \\ =\mathstrut \amp \left( -\frac{\pi}{2} \cos(\frac{\pi}{2}) + \sin(\frac{\pi}{2}) \right) - \left( -0 \cos(0) + \sin(0) \right) \\ =\mathstrut \amp 1 \text{.} \end{align*}

2. We can take \(u=\frac 1 2 t\) and \(v=e^{2t}\text{,}\) so that \(du =\frac 1 2 dt\) and \(dv=2e^{2t}dt\text{.}\) Then

   \begin{align*} \int_1^2 u\, dv =\mathstrut\amp \left. uv \right |_{t=1}^{t=2} - \int_1^2 v \, du \\ \int_1^2 te^{2t} \, dt = \mathstrut \amp \left. (\frac 12 t)(e^{2t}) \right |_{t=1}^{t=2} - \int_1^2 (e^{2t}) (\frac 1 2 dt) \\ = \mathstrut \amp \left. (\frac 12 t)(e^{2t}) \right |_{t=1}^{t=2}- \left. \frac 14 (e^{2t}) \right |_{t=1}^{t=2}\\ = \mathstrut \amp ((\frac 12 (2)))(e^{2(2))})-(\frac 12 (1))(e^{2(1)}))- (\frac 14 (e^{2(2)})-\frac 14 (e^{2(1)}))\\ = \mathstrut \amp \frac 34 e^4 - \frac 1 4 e^2 \approx 39.1\text{.} \end{align*}