# Step By Step Calculus » 15.3 - Integration by Parts

Synopsis
A powerful method for evaluating integrals is integration by parts; it has the following form:

 \displaystyle { \textbf{(Integration by Parts)}\qquad \int f(x)g^{\prime }(x)dx=f(x)g(x)-\int f^{\prime}(x)g(x)dx }\displaystyle { \textbf{(Integration by Parts)}\qquad \int f(x)g^{\prime }(x)dx=f(x)g(x)-\int f^{\prime}(x)g(x)dx }

and letting u=g(x)u=g(x) and v=f(x)v=f(x) one can express it as \displaystyle \int vdu=uv-\int udv. \displaystyle \int vdu=uv-\int udv.
Like the method of substitution, we still use the MESS procedure but obviously with a different interpretation for integration by parts.
 \displaystyle \textbf{(Manipulate) } \displaystyle \textbf{(Manipulate) } \displaystyle \displaystyle Identify f(x)f(x) and g'(x)g'(x) \displaystyle \textbf{(Eliminate) } \displaystyle \textbf{(Eliminate) } \displaystyle \displaystyle Write the integral as \displaystyle \int vdu \displaystyle \int vdu with \displaystyle v=f(x) \displaystyle v=f(x) and \displaystyle du=g'(x)dx\displaystyle du=g'(x)dx \displaystyle \textbf{(Solve) } \displaystyle \textbf{(Solve) } \displaystyle \int vdu =uv-\int udv \quad\displaystyle \int vdu =uv-\int udv \quad where \displaystyle u=\int g'(x)dx, dv=f'(x)dx\displaystyle u=\int g'(x)dx, dv=f'(x)dx \displaystyle \textbf{(Substitute) } \displaystyle \textbf{(Substitute) } \displaystyle \displaystyle Substitute back to an expression in terms of xx\displaystyle \displaystyle
Below are some common reduction formulae.
\begin{tabular}{l}\hline $\displaystyle I_n=\int \sin^n x dx=-\frac{1}{n}\sin ^{n-1}x\cos x+\frac{n-1}{n}I_{n-2},\; n\geq 2$ and $n\in\mathbb{Z^+}$ \\ Base Cases: $\displaystyle I_0=\int dx=x+c,\; I_1=\int\sin x dx=-\cos x+c$\\\hline $\displaystyle I_n=\int \cos^n x dx=\frac{1}{n}\cos ^{n-1}x\sin x+\frac{n-1}{n}I_{n-2},\; n\geq 2$ and $n\in\mathbb{Z^+}$ \\ Base Cases: $\displaystyle I_0=\int dx=x+c,\; I_1=\int\cos x dx=\sin x+c$\\\hline $\displaystyle I_n=\int \tan^n x dx=\frac{1}{n-1}\tan ^{n-1}x-I_{n-2},\; n\geq 2$ and $n\in\mathbb{Z^+}$ \\ Base Cases: $\displaystyle I_0=\int dx=x+c,\; I_1=\int\tan x dx=\ln |\sec x|+c$\\\hline $\displaystyle I_n=\int \cot^n x dx=-\frac{1}{n-1}\cot ^{n-1}x-I_{n-2},\; n\geq 2$ and $n\in\mathbb{Z^+}$ \\ Base Cases: $\displaystyle I_0=\int dx=x+c,\; I_1=\int\cot x dx=\ln |\sin x|+c$\\\hline $\displaystyle I_n=\int \sec^n x dx=\frac{1}{n-1}\sec ^{n-2}x\tan x+\frac{n-2}{n-1}I_{n-2},\; n\geq 2$ and $n\in\mathbb{Z^+}$ \\ Base Cases: $\displaystyle I_0=\int dx=x+c,\; I_1=\int\sec x dx=\ln |\sec x+\tan x|+c$\\\hline $\displaystyle I_n=\int \csc^n x dx=-\frac{1}{n-1}\csc ^{n-2}x\cot x+\frac{n-2}{n-1}I_{n-2},\; n\geq 2$ and $n\in\mathbb{Z^+}$ \\ Base Cases: $\displaystyle I_0=\int dx=x+c,\; I_1=\int\csc x dx=-\ln |\csc x+\cot x|+c$\\\hline $\displaystyle I_{(m,n)}=\int \sin^m x\cos^n dx=-\frac{\sin^{m-1}x\cos^{n+1}x}{m+n}+\frac{m-1}{m+n}I_{(m-2,n)},\; m\geq 2$ and $m\in\mathbb{Z^+}, n\in\mathbb{Z}$ \\ When $m=-n$, evaluate it as $\displaystyle \int\tan^m xdx$\\\hline $\displaystyle I_{(m,n)}=\int \sin^m x\cos^n dx=\frac{\sin^{m+1}x\cos^{n-1}x}{m+n}+\frac{n-1}{m+n}I_{(m,n-2)},\; n\geq 2$ and $n\in\mathbb{Z^+}, m\in\mathbb{Z}$ \\ When $m=-n$, evaluate it as $\displaystyle \int \cot^n xdx$\\\hline \end{tabular}
Evaluating Fractional Power Integrals: Given that \displaystyle (ax+b)^{\frac{1}{n}}\displaystyle (ax+b)^{\frac{1}{n}} makes sense, (i.e. either nn is odd or ax+bax+b is non-negative), we use the following steps to evaluate \displaystyle\int f\left( \left( ax+b\right) ^{\frac{1}{n}}\right) dx\displaystyle\int f\left( \left( ax+b\right) ^{\frac{1}{n}}\right) dx for n\in\mathbb{N}n\in\mathbb{N} with a,b\in\mathbb{R}a,b\in\mathbb{R}.
Step 1: (Substitute) We can make the substitution \displaystyle ax+b=t^{n}\displaystyle ax+b=t^{n}, \displaystyle dx=\frac{nt^{n-1}}{a}dt\displaystyle dx=\frac{nt^{n-1}}{a}dt so
 \displaystyle \int f\left( \left( ax+b\right) ^{\frac{1}{n}}\right) dx \displaystyle \int f\left( \left( ax+b\right) ^{\frac{1}{n}}\right) dx \displaystyle =\displaystyle = \displaystyle \left. \int f(t)\frac{nt^{n-1}}{a}dt\right| _{t=(ax+b)^{\frac{1}{n}}} \displaystyle \left. \int f(t)\frac{nt^{n-1}}{a}dt\right| _{t=(ax+b)^{\frac{1}{n}}} \displaystyle \displaystyle \displaystyle =\displaystyle = \displaystyle \frac{n}{a}\left. \int f(t)t^{n-1}dt\right| _{t=(ax+b)^{\frac{1}{n}}} \displaystyle \frac{n}{a}\left. \int f(t)t^{n-1}dt\right| _{t=(ax+b)^{\frac{1}{n}}}
Step 2: (Integrate by Parts) Use integration by parts to evaluate \displaystyle\int f(t)t^{n-1}dt\displaystyle\int f(t)t^{n-1}dt.