4. Integration by Parts

Recall:       \(\displaystyle \int u\,dv=u\,v-\int v\,du\)   where   \(du=\dfrac{du}{dx}\,dx\)   and   \(dv=\dfrac{dv}{dx}\,dx\)

d. Recurrence of the Integral

2. Reduction Formulas

Integration by parts can also be used to derive formulas which can be applied repeatedly to simplify an integral. The derivation again involves the recurrence of the original integral.

\[ \int \cos^n x\,dx =\dfrac{1}{n}\cos^{n-1}x\sin x+\dfrac{n-1}{n}\int \cos^{n-2}x\,dx \] (Do not memorize the reduction formulas.)

We first do integration by parts with \[\begin{array}{ll} u=\cos^{n-1}x & dv=\cos x\,dx \\ du=-(n-1)\cos^{n-2}x\sin x\,dx \quad & v=\sin x \end{array}\] So: \[ \int \cos^n x\,dx =\cos^{n-1}x\sin x+\int (n-1)\cos^{n-2}x\sin^2 x\,dx \] Within the remaining integral, we use the identity \(\sin^2 x=1-\cos^2 x\), multiply out the integrand and write the remaining integral as a sum of two integrals:: \[\begin{aligned} \int \cos^n x\,dx &=\cos^{n-1}x\sin x+(n-1)\int \cos^{n-2}x(1-\cos^2 x)\,dx \\ &=\cos^{n-1}x\sin x+(n-1)\int \cos^{n-2}x-\cos^n x\,dx \\ &=\cos^{n-1}x\sin x+(n-1)\int \cos^{n-2}x\,dx-(n-1)\int \cos^n x\,dx \end{aligned}\] Once again, the original integral recurs on the right. So we solve for it: \[\begin{aligned} n\int \cos^n x\,dx &=\cos^{n-1}x\sin x+(n-1)\int \cos^{n-2}x\,dx \\ \int \cos^n x\,dx &=\dfrac{1}{n}\cos^{n-1}x\sin x+\dfrac{n-1}{n}\int \cos^{n-2}x\,dx \end{aligned}\] which is the desired reduction formula. Note, we do not need a \(+C\) because there is still an unevaluated integral on the right.

Use the reduction formula \[ \int \cos^n x\,dx =\dfrac{1}{n}\cos^{n-1}x\sin x+\dfrac{n-1}{n}\int \cos^{n-2}x\,dx \] to compute \(\displaystyle \int \cos^2 x\,dx\) and \(\displaystyle \int \cos^4 x\,dx\).

We first use the reduction formula with \(n=2\): \[ \int \cos^2 x\,dx =\dfrac{1}{2}\cos^1 x\sin x+\dfrac{1}{2}\int 1\,dx \] To complete this, we do the integral: \[ \int \cos^2 x\,dx =\dfrac{\sin x\cos x}{2}+\dfrac{x}{2}+C \] Next, we use the reduction formula with \(n=4\): \[ \int \cos^4 x\,dx =\dfrac{1}{4}\cos^3 x\sin x+\dfrac{3}{4}\int \cos^2 x\,dx \] and substitute in the formula for \(\displaystyle \int \cos^2 x\,dx\): \[\begin{aligned} \int \cos^4 x\,dx &=\dfrac{1}{4}\cos^3 x\sin x +\dfrac{3}{4}\left(\dfrac{\sin x\cos x}{2}+\dfrac{x}{2}\right)+C \\ &=\dfrac{\cos^3 x\sin x}{4}+\dfrac{3\sin x\cos x}{8}+\dfrac{3x}{8}+C \end{aligned}\] Of course you could also do either integral by using the trig identity \(\cos^2 x=\dfrac{1+\cos 2x}{2}\). We will cover that method in more detail in the chapter on Trig Integrals.