4. Integration by Parts
Recall: \(\displaystyle \int u\,dv=u\,v-\int v\,du\) where \(du=\dfrac{du}{dx}\,dx\) and \(dv=\dfrac{dv}{dx}\,dx\)
b2. Integrals of Special Functions
Integration by parts can be used to find the integrals of many additional special functions.
Compute \(\displaystyle \int \arcsin x\,dx\).
We do not appear to have the product of two functions here. However, one of the functions can be taken as \(1\). So we take \[\begin{array}{ll} u=\arcsin x & dv=1\,dx=dx \\ du=\dfrac{1}{\sqrt{1-x^2}}\,dx \quad & v=x \end{array}\] Thus \[ \int \arcsin x\,dx =x\arcsin x-\int \dfrac{x}{\sqrt{1-x^2}}\,dx \] To compute the remaining integral, we make the substitution \(u=1-x^2\). Then \(du=-2x\,dx\) and so \(x\,dx=-\,\dfrac{1}{2}\,du\). Thus, \[\begin{aligned} \int \arcsin x\,dx &=x\arcsin x+\dfrac{1}{2}\int \dfrac{1}{\sqrt{u}}\,du \\ &=x\arcsin x+\sqrt{u}+C \\ &=x\arcsin x+\sqrt{1-x^2}+C \end{aligned}\]
We check by differentiating: If \(f(x)=x\arcsin x+\sqrt{1-x^2}\), then \[ f'(x)=\arcsin x+x\dfrac{1}{\sqrt{1-x^2}} +\dfrac{1}{2}\dfrac{-2x}{\sqrt{1-x^2}}=\arcsin x \] which is the original integrand.
To compute \(\displaystyle \int \arccos x\,dx\), what 'parts' should be selected?
\(\displaystyle\begin{aligned}&u=\arccos x \\ &dv=dx\end{aligned}\)
\(\displaystyle\begin{aligned}&u=\arccos x \\ &dv=\cos x\,dx\end{aligned}\)
\(\displaystyle\begin{aligned}&u=1 \\ &dv=\arccos x\,dx\end{aligned}\)
\(\displaystyle\begin{aligned}&u=\cos x \\ &dv=dx\end{aligned}\)
A. Correct. This selection of parts gives \[ du=\dfrac{-1}{\sqrt{1-x^2}} \qquad \text{and} \qquad v=x \] This gives \[ \int \arccos x\,dx=x\arccos x+\int \dfrac{x}{\sqrt{1-x^2}}dx \] The problem is now very much like the one worked out above. Use the substitution \(u=1-x^2\) to finish the problem.
B. Sorry. You seem to have separated the \(\arccos\) function into two parts. The expression \(\arccos\) denotes a function --- and cannot be grammatically split even if both parts make mathematical sense. (In this case they don't.) Try again.
C. Incorrect. Resolving the problem this way involves finding \(v\) and that is the problem we started with.
D. Sorry. This problem involves the \(\arccos\) function, which is the inverse of the \(\cos\) function. Try again.
Compute \(\displaystyle \int \ln x\,dx\).
Take \(u=\ln x\) and \(dv=dx\).
\(\displaystyle \int \ln x\,dx=x\ln x-x+C\).
Take \[\begin{array}{ll} u=\ln x & dv=dx \\ du=\dfrac{1}{x}\,dx \quad & v=x \end{array}\] Then \[ \int \ln x\,dx=x\ln x-\int x\dfrac{1}{x}\,dx=x\ln x-x+C \]
We check using the Product Rule: \[\begin{aligned} \dfrac{d}{dx}&(x\ln x-x) \\ &=\ln x+x\dfrac{1}{x}-1=\ln x \end{aligned}\]
Compute\(\displaystyle \int \arctan x\,dx\).
Take \(u=\arctan x\) and \(dv=dx\).
\(\displaystyle \int \arctan x\,dx =x\arctan x-\dfrac{1}{2}\ln(1+x^2)+C\).
Take \[\begin{array}{ll} u=\arctan x & dv=dx \\ du=\dfrac{1}{1+x^2}\,dx \quad & v=x \end{array}\] Then \[ \int \arctan x\,dx=x\arctan x-\int \dfrac{x}{1+x^2}\,dx \] Now make the substitution \(u=1+x^2\). Then \(du=2x\,dx\). So \[\begin{aligned} \int \arctan x\,dx &=x\arctan x-\dfrac{1}{2}\int \dfrac{1}{u}\,du\ \\ &=x\arctan x-\dfrac{1}{2}\ln|u|+C\ \\ &=x\arctan x-\dfrac{1}{2}\ln(1+x^2)+C \end{aligned}\]
We check using the Product Rule: \[\begin{aligned} \dfrac{d}{dx}&\left(x\arctan x-\dfrac{1}{2}\ln(1+x^2)\right) \\ &=\arctan x+\dfrac{x}{1+x^2}-\dfrac{1}{2}\dfrac{2x}{1+x^2} =\arctan x \end{aligned}\]
Compute\(\displaystyle \int \text{arcsec}\,x\,dx\).
(Look at \(x \ge 1\) and \(x \le -1\) separately.)
This one is really tricky. Read the hint or just read the solution.
First, recall that \(\text{arcsec}\,x \) is only defined for \(|x| \ge 1\)
because \(|\sec\theta| \ge 1\) because \(|\cos\theta| \le 1\). So you need
to do \(x \ge 1\) and \(x \le -1\) separately.
Second, the solution uses the derivative:
\[
\dfrac{d}{dx}\text{arcsec}\,x=\dfrac{1}{|x|\sqrt{x^2-1}}
\]
and the integral:
\[
\int \sec\theta\,d\theta=\ln|\sec\theta+\tan\theta|+C
\]
This integral will be derived in the chapter on
Trig Integrals. However, it can be easily checked by differentiating:
\[\begin{aligned}
\dfrac{d}{d\theta}\ln|\sec\theta+\tan\theta|
&=\dfrac{1}{\sec\theta+\tan\theta}(\sec\theta\tan\theta+\sec^2\theta) \\[5pt]
&=\dfrac{\sec\theta(\tan\theta+\sec\theta)}{\sec\theta+\tan\theta}
=\sec\theta
\end{aligned}\]
Third, to do the integration by parts, take \(u=\text{arcsec}\,x\) and \(dv=dx\).
Then make the substitution \(x=\sec\theta\) and use the Pythagorean
identity for \(\tan\) and \(\sec\).
\(\displaystyle \int \text{arcsec}\,x\,dx =x\,\text{arcsec}\,x-\ln\left(|x|+\sqrt{x^2-1}\right)+C \)
We do integration by parts with: \[\begin{array}{ll} u=\text{arcsec}\,x & dv=dx \\ du=\dfrac{1}{|x|\sqrt{x^2-1}}\,dx \quad & v=x \end{array}\] This gives: \[ \int \text{arcsec}\,x\,dx=x\,\text{arcsec}\,x-\int \dfrac{x}{|x|\sqrt{x^2-1}}\,dx \]
For \(x \ge 1\), this becomes: \[ \int \text{arcsec}\,x\,dx=x\,\text{arcsec}\,x-\int \dfrac{1}{\sqrt{x^2-1}}\,dx \] We make the substitution \(x=\sec\theta\) with \(0 \le \theta \lt \dfrac{\pi}{2}\). Then \(\sec\theta \ge 1\) and \(\tan\theta \ge 0\). To make the substitution, we use \(dx=\sec\theta\tan\theta\,d\theta\) and \(x^2-1=\sec^2\theta-1=\tan^2\theta\). We get: \[\begin{aligned} \int \text{arcsec}\,x\,dx &=x\,\text{arcsec}\,x-\int \dfrac{\sec\theta\tan\theta}{\sqrt{\tan^2\theta}}\,d\theta \\ &=x\,\text{arcsec}\,x-\int \sec\theta\,d\theta \\ &=x\,\text{arcsec}\,x-\ln|\sec\theta+\tan\theta|+C \end{aligned}\] To substitute back, we have \(\sec\theta=x \ge 1\) and \(\tan\theta=\sqrt{x^2-1} \ge 0\). So: \[\begin{aligned} \int \text{arcsec}\,x\,dx &=x\,\text{arcsec}\,x-\ln\left(x+\sqrt{x^2-1}\right)+C \end{aligned}\]
On the other hand, for \(x \le -1\), the integration by parts formula becomes: \[ \int \text{arcsec}\,x\,dx =x\,\text{arcsec}\,x+\int \dfrac{1}{\sqrt{x^2-1}}\,dx \] We now make the substitution \(x=-\sec\theta\) with \(0 \le \theta \lt \dfrac{\pi}{2}\). Then \(\sec\theta \ge 1\) and \(\tan\theta \ge 0\). To make the substitution, we use \(dx=-\sec\theta\tan\theta\,d\theta\) and \(x^2-1=\sec^2\theta-1=\tan^2\theta\). So: \[\begin{aligned} \int \text{arcsec}\,x\,dx &=x\,\text{arcsec}\,x-\int \dfrac{\sec\theta\tan\theta}{\sqrt{\tan^2\theta}}\,d\theta \\ &=x\,\text{arcsec}\,x-\int \sec\theta\,d\theta \\ &=x\,\text{arcsec}\,x-\ln|\sec\theta+\tan\theta|+C \end{aligned}\] To substitute back, we have \(\sec\theta=-x \ge 1\) and \(\tan\theta=\sqrt{x^2-1} \ge 0\). Therefore: \[ \int \text{arcsec}\,x\,dx =x\,\text{arcsec}\,x-\ln\left(-x+\sqrt{x^2-1}\right)+C \]
Together they say: \[ \int \text{arcsec}\,x\,dx =x\,\text{arcsec}\,x-\ln\left(|x|+\sqrt{x^2-1}\right)+C \]
We check using the Product Rule and the Chain Rule. For \(x \ge 1\): \[\begin{aligned} \dfrac{d}{dx}&\left[x\,\text{arcsec}\,x-\ln\left(x+\sqrt{x^2-1}\right)\right] \\ &=\text{arcsec}\,x+\dfrac{x}{|x|\sqrt{x^2-1}} -\dfrac{1}{x+\sqrt{x^2-1}}\left(1+\dfrac{x}{\sqrt{x^2-1}}\right)\\ &=\text{arcsec}\,x+\dfrac{x}{x\sqrt{x^2-1}} -\dfrac{1}{x+\sqrt{x^2-1}}\dfrac{\sqrt{x^2-1}+x}{\sqrt{x^2-1}}\\ &=\text{arcsec}\,x+\dfrac{1}{\sqrt{x^2-1}}-\dfrac{1}{\sqrt{x^2-1}} \\ &=\text{arcsec}\,x \end{aligned}\] For \(x \le -1\): \[\begin{aligned} \dfrac{d}{dx}&\left[x\,\text{arcsec}\,x-\ln\left(-x+\sqrt{x^2-1}\right)\right] \\ &=\text{arcsec}\,x+\dfrac{x}{|x|\sqrt{x^2-1}} -\dfrac{1}{-x+\sqrt{x^2-1}}\left(-1+\dfrac{x}{\sqrt{x^2-1}}\right)\\ &=\text{arcsec}\,x+\dfrac{x}{-x\sqrt{x^2-1}} -\dfrac{1}{-x+\sqrt{x^2-1}}\dfrac{-\sqrt{x^2-1}+x}{\sqrt{x^2-1}}\\ &=\text{arcsec}\,x-\dfrac{1}{\sqrt{x^2-1}}+\dfrac{1}{\sqrt{x^2-1}} \\ &=\text{arcsec}\,x \end{aligned}\]
As a result of these computations, we now know:
\[\begin{aligned} &\int \ln x\,dx=x\ln x-x+C \\ &\int \arcsin x\,dx=x\arcsin x+\sqrt{1-x^2}+C \\ &\int \arctan x\,dx=x\arctan x-\dfrac{1}{2}\ln(1+x^2)+C \\ &\int \text{arcsec}\,x\,dx =x\,\text{arcsec}\,x-\ln\left(|x|+\sqrt{x^2-1}\right)+C \end{aligned}\] Only \(\displaystyle \int \ln x\,dx\) is worth memorizing.
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