4. Integration by Parts
Recall: \(\displaystyle \int u\,dv=u\,v-\int v\,du\) where \(du=\dfrac{du}{dx}\,dx\) and \(dv=\dfrac{dv}{dx}\,dx\)
b1. Choosing the 'Parts'
c. Inverse Trig Examples
Compute \(\displaystyle \int x^2\arctan x\,dx\).
Since \(\arctan\) is hard to integrate, we start with the parts: \[\begin{array}{ll} u=\arctan x & dv=x^2\,dx \\ du=\dfrac{1}{1+x^2}\,dx \quad & v=\dfrac{1}{3}x^3 \end{array}\] This is the “A” in LAPTE. This gives: \[ \int x^2\arctan x\,dx =\dfrac{1}{3}x^3\arctan x-\dfrac{1}{3}\int \dfrac{x^3}{1+x^2}\,dx \] For the remaining integral, we apply polynomial long division, (or simply ) to get:
\[ \dfrac{x^3}{1+x^2} =\dfrac{x^3+x-x}{1+x^2} =\dfrac{x(x^2+1)}{1+x^2}-\dfrac{x}{1+x^2} =x-\dfrac{x}{1+x^2} \]
\[\begin{aligned} \int x^2\arctan x\,dx &=\dfrac{1}{3}x^3\arctan x-\dfrac{1}{3}\int \left(x-\dfrac{x}{1+x^2}\right)\,dx \\ &=\dfrac{1}{3}x^3\arctan x-\dfrac{1}{3}\int x\,dx+\dfrac{1}{3}\int \dfrac{x}{1+x^2}\,dx \end{aligned}\] The last two integrals are much easier: \[ \int x^2\arctan x\,dx= \dfrac{1}{3}x^3\arctan x-\dfrac{1}{6}x^2+\dfrac{1}{6}\ln(1+x^2)+C \]
We check using the Product Rule: \[\begin{aligned} \dfrac{d}{dx}&\left(\dfrac{1}{3}x^3\arctan x-\dfrac{1}{6}x^2+\dfrac{1}{6}\ln(1+x^2)\right) \\ &=x^2\arctan x+\dfrac{x^3}{3}\dfrac{1}{1+x^2} -\dfrac{1}{3}x+\dfrac{1}{3}\dfrac{x}{1+x^2} \\ &=x^2\arctan x+\dfrac{1}{3}\dfrac{x^3-x(1+x^2)+x}{1+x^2} \\ &=x^2\arctan x \end{aligned}\]
Now your turn:
Compute \(\displaystyle \int x^2\arcsin x\,dx\).
\(\displaystyle \int x^2\arcsin x\,dx =\dfrac{x^3}{3}\arcsin x+\dfrac{1}{9}(x^2+2)\sqrt{1-x^2}+C\)
Since \(\arcsin\) is hard to integrate, we start with the parts: \[\begin{array}{ll} u=\arcsin x & dv=x^2\,dx \\ du=\dfrac{1}{\sqrt{1-x^2}}\,dx \quad & v=\dfrac{x^3}{3} \end{array}\] This is again the “A” in LAPTE. That gives us: \[ \int x^2\arcsin x\,dx =\dfrac{x^3}{3}\arcsin x-\dfrac{1}{3}\int \dfrac{x^3}{\sqrt{1-x^2}}\,dx \] From here we use a substitution, but to avoid confusion with the integration by parts, we'll use \(z\) instead of \(u\). We set \[\begin{aligned} z=1-x^2 \qquad &\text{and} \qquad dz=-2x\,dx \\ x^2=1-z \qquad &\text{and} \qquad -\,\dfrac{1}{2}\,dz=x\,dx \end{aligned}\] Then: \[\begin{aligned} \int x^2\arcsin x\,dx &=\dfrac{x^3}{3}\arcsin x+\dfrac{1}{6}\int \dfrac{1-z}{\sqrt{z}}\,dz \\ &=\dfrac{x^3}{3}\arcsin x+\dfrac{1}{6}\int \dfrac{1}{\sqrt{z}}\,dz -\dfrac{1}{6}\int \sqrt{z}\,dz \\ &=\dfrac{x^3}{3}\arcsin x+\dfrac{1}{3}\sqrt{z}-\dfrac{1}{9}z^{3/2}+C \\ &=\dfrac{x^3}{3}\arcsin x+\dfrac{1}{3}\sqrt{1-x^2}-\dfrac{1}{9}(1-x^2)^{3/2}+C \\ &=\dfrac{x^3}{3}\arcsin x+\dfrac{1}{9}(x^2+2)\sqrt{1-x^2}+C \end{aligned}\]
We check using the Product Rule: \[\begin{aligned} \dfrac{d}{dx}&\left(\dfrac{x^3}{3}\arcsin x+\dfrac{1}{9}(x^2+2)\sqrt{1-x^2}\right) \\ &=x^2\arcsin x+\dfrac{x^3}{3}\dfrac{1}{\sqrt{1-x^2}} +\dfrac{1}{9}(2x)\sqrt{1-x^2}+\dfrac{1}{9}(x^2+2)\dfrac{-x}{\sqrt{1-x^2}} \\ &=x^2\arcsin x+\dfrac{3x^3+2x(1-x^2)-x(x^2+2)}{9\sqrt{1-x^2}} \\ &=x^2\arcsin x \end{aligned}\]