4. Integration by Parts

Recall:       ∫udv=uv−∫vdu\displaystyle \int u\,dv=u\,v-\int v\,du   where   du=dudxdxdu=\dfrac{du}{dx}\,dx   and   dv=dvdxdxdv=\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.

Derive the reduction formula
∫cos⁡nxdx=1ncos⁡n−1xsin⁡x+n−1n∫cos⁡n−2xdx \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 u=cos⁡n−1xdv=cos⁡xdxdu=−(n−1)cos⁡n−2xsin⁡xdxv=sin⁡x\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: ∫cos⁡nxdx=cos⁡n−1xsin⁡x+∫(n−1)cos⁡n−2xsin⁡2xdx \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⁡2x=1−cos⁡2x\sin^2 x=1-\cos^2 x, multiply out the integrand and write the remaining integral as a sum of two integrals:: ∫cos⁡nxdx=cos⁡n−1xsin⁡x+(n−1)∫cos⁡n−2x(1−cos⁡2x)dx=cos⁡n−1xsin⁡x+(n−1)∫cos⁡n−2x−cos⁡nxdx=cos⁡n−1xsin⁡x+(n−1)∫cos⁡n−2xdx−(n−1)∫cos⁡nxdx\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: n∫cos⁡nxdx=cos⁡n−1xsin⁡x+(n−1)∫cos⁡n−2xdx∫cos⁡nxdx=1ncos⁡n−1xsin⁡x+n−1n∫cos⁡n−2xdx\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+C because there is still an unevaluated integral on the right.

Use the reduction formula ∫cos⁡nxdx=1ncos⁡n−1xsin⁡x+n−1n∫cos⁡n−2xdx \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 ∫cos⁡2xdx\displaystyle \int \cos^2 x\,dx and ∫cos⁡4xdx\displaystyle \int \cos^4 x\,dx.

We first use the reduction formula with n=2n=2: ∫cos⁡2xdx=12cos⁡1xsin⁡x+12∫1dx \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: ∫cos⁡2xdx=sin⁡xcos⁡x2+x2+C \int \cos^2 x\,dx =\dfrac{\sin x\cos x}{2}+\dfrac{x}{2}+C Next, we use the reduction formula with n=4n=4: ∫cos⁡4xdx=14cos⁡3xsin⁡x+34∫cos⁡2xdx \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 ∫cos⁡2xdx\displaystyle \int \cos^2 x\,dx: ∫cos⁡4xdx=14cos⁡3xsin⁡x+34(sin⁡xcos⁡x2+x2)+C=cos⁡3xsin⁡x4+3sin⁡xcos⁡x8+3x8+C\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⁡2x=1+cos⁡2x2\cos^2 x=\dfrac{1+\cos 2x}{2}. We will cover that method in more detail in the chapter on Trig Integrals.

Derive a reduction formula for ∫sin⁡nxdx\displaystyle \int \sin^n x\,dx

Answer

∫sin⁡nxdx=−1nsin⁡n−1xcos⁡x+n−1n∫sin⁡n−2xdx\displaystyle \int \sin^n x\,dx =-\,\dfrac{1}{n}\sin^{n-1}x\cos x+\dfrac{n-1}{n}\int \sin^{n-2}x\,dx

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Solution

We first do integration by parts with u=sin⁡n−1xdv=sin⁡xdxdu=(n−1)sin⁡n−2xcos⁡xdxv=−cos⁡x\begin{array}{ll} u=\sin^{n-1}x & dv=\sin x\,dx \\ du=(n-1)\sin^{n-2}x\cos x\,dx \quad & v=-\cos x \end{array} So: ∫sin⁡nxdx=−sin⁡n−1xcos⁡x+∫(n−1)sin⁡n−2xcos⁡2xdx \int \sin^n x\,dx =-\sin^{n-1}x\cos x+\int (n-1)\sin^{n-2}x\cos^2 x\,dx Within the resulting integral, we use the identity cos⁡2x=1−sin⁡2x\cos^2 x=1-\sin^2 x, multiply out the integrand and write the resulting integral as a sum of two integrals:: ∫sin⁡nxdx=−sin⁡n−1xcos⁡x+(n−1)∫sin⁡n−2x(1−sin⁡2x)dx=−sin⁡n−1xcos⁡x+(n−1)∫sin⁡n−2x−sin⁡nxdx=−sin⁡n−1xcos⁡x+(n−1)∫sin⁡n−2xdx−(n−1)∫sin⁡nxdx\begin{aligned} \int \sin^n x\,dx &=-\sin^{n-1}x\cos x+(n-1)\int \sin^{n-2}x(1-\sin^2 x)\,dx \\ &=-\sin^{n-1}x\cos x+(n-1)\int \sin^{n-2}x-\sin^n x\,dx \\ &=-\sin^{n-1}x\cos x+(n-1)\int \sin^{n-2}x\,dx-(n-1)\int \sin^n x\,dx \end{aligned} Once again, the original integral recurs on the right. So we solve for it: n∫sin⁡nxdx=−sin⁡n−1xcos⁡x+(n−1)∫sin⁡n−2xdx∫sin⁡nxdx=−1nsin⁡n−1xcos⁡x+n−1n∫sin⁡n−2xdx\begin{aligned} n\int \sin^n x\,dx &=-\sin^{n-1}x\cos x+(n-1)\int \sin^{n-2}x\,dx \\ \int \sin^n x\,dx &=-\,\dfrac{1}{n}\sin^{n-1}x\cos x+\dfrac{n-1}{n}\int \sin^{n-2}x\,dx \end{aligned} which is the desired reduction formula.

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Using this reduction formula, compute the integral ∫sin⁡3xdx\displaystyle \int \sin^3 x\,dx.

Answer

∫sin⁡3xdx=−13sin⁡2xcos⁡x−23cos⁡x+C\displaystyle \int \sin^3 x\,dx =-\,\dfrac{1}{3}\sin^2 x\cos x-\dfrac{2}{3}\cos x+C

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Solution

Let's start with the reduction formula: ∫sin⁡nxdx=−1nsin⁡n−1xcos⁡x+n−1n∫sin⁡n−2xdx \int \sin^n x\,dx =-\,\dfrac{1}{n}\sin^{n-1}x\cos x+\dfrac{n-1}{n}\int \sin^{n-2}x\,dx Plugging in n=3n=3, we get ∫sin⁡3xdx=−13sin⁡2xcos⁡x+23∫sin⁡xdx=−13sin⁡2xcos⁡x−23cos⁡x+C\begin{aligned} \int \sin^{3} x\,dx &=-\,\dfrac{1}{3}\sin^2 x\cos x+\dfrac{2}{3}\int \sin x\,dx \\ &=-\,\dfrac{1}{3}\sin^2 x\cos x-\dfrac{2}{3}\cos x+C \end{aligned}

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Check

We check by differentiating. If f=−13sin⁡2xcos⁡x−23cos⁡xf=-\,\dfrac{1}{3}\sin^2 x\cos x-\dfrac{2}{3}\cos x, then f′=−23sin⁡xcos⁡2x+13sin⁡3x+23sin⁡x=−23sin⁡x(1−sin⁡2x)+13sin⁡3x+23sin⁡x=sin⁡3x\begin{aligned} f'&=-\,\dfrac{2}{3}\sin x\cos^2 x+\dfrac{1}{3}\sin^3 x+\dfrac{2}{3}\sin x \\ &=-\,\dfrac{2}{3}\sin x(1-\sin^2 x)+\dfrac{1}{3}\sin^3 x+\dfrac{2}{3}\sin x \\ &=\sin^3 x \end{aligned}

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Remark

The integral can also be done with the substitution u=cos⁡xu=\cos x and du=−sin⁡xdxdu=-\sin x\,dx: ∫sin⁡3xdx=∫(1−cos⁡2x)sin⁡xdx=−∫(1−u2)du=−u+u33+C=−cos⁡x+cos⁡3x3+C\begin{aligned} \int \sin^3 x\,dx &=\int (1-\cos^2 x)\sin x\,dx =-\int (1-u^2)\,du \\ &=-u+\dfrac{u^3}{3}+C =-\cos x+\dfrac{\cos^3 x}{3}+C \end{aligned} We check this is equivalent to the previous solution: −13sin⁡2xcos⁡x−23cos⁡x+C=−13(1−cos⁡2x)cos⁡x−23cos⁡x+C=−cos⁡x+cos⁡3x3+C\begin{aligned} -\,\dfrac{1}{3}\sin^2 x\cos x-\dfrac{2}{3}\cos x+C &=-\,\dfrac{1}{3}(1-\cos^2 x)\cos x-\dfrac{2}{3}\cos x+C \\ &=-\cos x+\dfrac{\cos^3 x}{3}+C \end{aligned}

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