Integration by parts

The integration by parts applies if the integrand consists two parts (functions) in which one function is NOT the derivative of other. This rule is a special rule based on product rule for derivatives:
\( \left( uv \right)'=uv'+u'v\)
Integrate both sides and rearrange, we get
\( \int \left( uv \right)'dx=\int uv'dx+\int u'vdx\)
or \( uv=\int uv'dx+\int u'vdx\)
or \( \int uv'dx=uv-\int u'vdx\)
Integrate \( v\) as a function, so the left side remain in product form
\( \int (uv) dx=u\int vdx-\int u'\left( \int vdx \right)dx\) (1)

To identify the first function (u) and second function (v), we Follow ILATE principle.

1. I: Inverse trigonometric functions such as \( \sin^{-1}(x), \cos ^{-1}(x), \tan ^{-1}(x)\)
2. L: Logarithmic functions such as \(log(x), log(x)\)
3. A: Algebraic functions such as \(x^2, x^3\)
4. T: Trigonometric functions such as \(\sin (x), \cos (x), \tan (x)\)
5. E: Exponential functions such as \(e^x, 3^x\)

The ILATE principle is based on the easyness of derivative and integration.

1. The derivative is easier when function comes from Top to Down (I → L → A → T → E)
2. The integral is easier when function goes from Down to Top (E → T → A → L → I)
3. If both function are of same type, we choose first function the one, wwhose derivative is easier (vanishes at some order), or we choose second function the one, wwhose integral is easier.
4. Is ILATE principle compoulsary? NO, we can solve without ILATE principle as well, if applicable

Solved Examples

Evaluate the following integrals

1. \( \int x^2 \sin x dx\)
   
   Solution
   By ILATE rule, we let
   \( u=x^2, v=\sin x\) then du=2x, and \( \int v dx =-\ cos x\)
   Now, the integral is
   \( \int x^2 \sin x dx\)
   or\( \int uv dx \)
   or\( u \int v- \int (du) \int v \)
   or\( x^2 (- \cos x)- \int (2x) (-\cos x) dx\)
   or\( -x^2 \cos x+ \int 2x \cos x dx\)(A)
   Again, we do integration for \( \int 2x \cos x dx\)
   Here, we let let
   u=2x, v=cos x then du=2, and \( \int v dx =\sin x\)
   Now
   \( \int 2x \cos x dx\)
   or\( \int uv dx \)
   or\( u \int v- \int (du) v \)
   or\( 2x (\sin x)- \int (2) (\sin x)dx\)
   or\( 2x \sin x-2 (-\cos x)+c\)
   or\( 2x \sin x+2\cos x+c\)(B)
   Using (B) in (A), the answer is
   \( -x^2 \cos x+ \int 2x \cos x dx\)
   or\( -x^2 \cos x+ 2x \sin x+2\cos x+c\)
2. \( \int x^2 \sin x dx\)
   
   Solution
   By ILATE rule, we let
   \( u=x^2, v=\sin x\), then we prepare the table as below
   

   Sign	+	-	+	-
   Successive Derivative of u, Start from u	x2	2x	2	0
   Successive Integration of v, Start from \( \int v\)	-cos x	-sin x	cos x	sin x
   Product	\( -x^2\cos x\)	\( +2x \sin x\)	\( 2 \cos x\)	0

   Hence, the answer is
   \( \int x^2 \sin x dx=-x^2\cos x+2x \sin x+2 \cos x\)
3. \( \int x^2 \sin ^{-1} xdx\)
   
   Solution
   By ILATE rule, we let \(u=\sin ^{-1} x, v=x^2\), in which \(u'=\frac{1}{\sqrt{1-x^2}}, \int v=\frac{x^3}{3}\)
   Hence
   \(I=\int uv dx\)
   or \(I=u \int v-\int (u' \int v)\)
   or \(I=\sin ^{-1} x \frac{x^3}{3} -\int \left ( \frac{1}{\sqrt{1-x^2}} \frac{x^3}{3} \right) dx\)
   or \(I=\sin ^{-1} x \frac{x^3}{3} -\int \left ( \frac{1}{\sqrt{1-x^2}} \frac{x^3}{3} \right) dx\)
   or \(I=\frac{1}{3} x^3 \sin ^{-1} x - \frac{1}{3} \int \frac{x^3}{\sqrt{1-x^2}} dx \)
   Substitute \(u=x^2\) then \(2xdx=du\),this gives that
   \(I=\frac{1}{3} x^3 \sin ^{-1} x - \frac{1}{6} \int \frac{u}{\sqrt{1-u}} du \)
   Now, we put s=1-u, then du=-ds, hence, we get
   \(I=\frac{1}{3} x^3 \sin ^{-1} x + \frac{1}{6} \int \frac{1-s}{\sqrt{s}} ds \)
   or \(I=\frac{1}{3} x^3 \sin ^{-1} x + \frac{1}{6} \int \left (\frac{1}{\sqrt{s}}-\sqrt{s} \right ) ds \)
   or \(I=\frac{1}{3} x^3 \sin ^{-1} x + \frac{\sqrt{s}}{3}-\frac{s^{3/2}}{9} \)
   Replacing s=1-u, we get
   \(I=\frac{1}{3} x^3 \sin ^{-1} x + \frac{\sqrt{1-u}}{3}-\frac{(1-u)^{3/2}}{9} \)
   Replacing \(u=x^2\), we get
   \(I=\frac{1}{3} x^3 \sin ^{-1} x + \frac{\sqrt{1-x^2}}{3}-\frac{(1-x^2)^{3/2}}{9} \)
4. \( \int \frac{\sqrt{x^2+1}[\log (x^2+1)-2\log x]}{x^4}dx\)
   \( \int \frac{\sqrt{x^2+1}\log (1+\frac{1}{x^2})}{x^4}dx\)
   \( \int \frac{\sqrt{x^2+1}}{x} \frac{\log (1+\frac{1}{x^2})}{x^3}dx\)
   Put \( 1+\frac{1}{x^2}=u\) then \(\frac{dx}{x^3}=-\frac{1}{2}du\)
5. \( \tan^{-1}\sqrt{x} dx\)
   Put \(\sqrt{x}=u\)

Exercise: Evaluate the following integrals

SN	Question	Answer
1	\( \int x^2\cos xdx\)	\( x^2\sin x-2\sin x+2x\cos x+c\)
2	\( \int x\cos xdx\)	\( \cos x+x\sin x+c\)
3	\( \int x^3\sin xdx\)	\( -x^3\cos x+3x^2\sin x+6x\cos x-6\sin x+c\)
4	\( \int xe^{x^2}dx \)	\( \frac{1}{2} e^{x^2}+c\)
5	\(\int \sec^2 x \csc^2 xdx \)	\( \sec x \csc x-2 \cot x+c\)
6	\( \int \log xdx\)	\( x \log x-x+c\)
7	\( \int x\tan^{-1} xdx\)	\( \frac{1}{2}(x^2\tan^{-1} x+\tan^{-1} x-x)+c\)
8	\( \int xe^xdx \)	\( (x-1)e^x+c\)
9	\( \int x^3\cos xdx\)	\( x^3\sin x+3x^2\cos x-6x\sin x-6\cos x+c\)
10	\( \int{x e^x}dx\)
11	\( \int x\sin x\cos xdx \)	\(\frac{x}{4}-\frac{x\cos^ 2x}{2}+\frac{\cos x\sin x}{4}+c\)
12	\( \int \sin^ 2xdx \)	\( \frac{x}{2} -\frac{\sin (2x)}{4}+c\)
13	\( \int x\sin^ 2xdx\)	\(\frac{x^2}{4}-\frac{\cos^ 2x}{4}-\frac{x\sin x\cos x}{2}+c\)