Integration by Parts

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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:
${\left(uv\right)}^{\prime }=u{v}^{\prime }+u^{\prime }v$
Integrate both sides and rearrange, we get
$\int {\left(uv\right)}^{\prime }dx=\int u{v}^{\prime }dx+\int u^{\prime }vdx$
or $uv=\int u{v}^{\prime }dx+\int u^{\prime }vdx$
or $\int u{v}^{\prime }dx=uv-\int u^{\prime }vdx$
Integrate $v$ as a function, so the left side remain in product form
$\int (uv)dx=u\int vdx-\int u^{\prime }(\int vdx)dx$
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 Right to Left
2. The integral is easier when function goes from Left to Right
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

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Solved Examples

Evaluate the following integrals

1. $\int x^2\sin xdx$

   Solution 👉 Click Here

   Solution
   By ILATE rule, we let
   $u=x^2,v=\sin x$ then du=2x, and $\int vdx=-cosx$
   Now, the integral is
   $\int x^2\sin xdx$
   or$\int uvdx$
   or$u\int v-\int (du)\int v$
   or$x^2(-\cos x)-\int (2x)(-\cos x)dx$
   or$-x^2\cos x+\int 2x\cos xdx$(A)
   Again, we do integration for $\int 2x\cos xdx$
   Here, we let let
   u=2x, v=cos x then du=2, and $\int vdx=\sin x$
   Now
   $\int 2x\cos xdx$
   or$\int uvdx$
   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
   $-x^2\cos x+\int 2x\cos xdx$
   or$-x^2\cos x+2x\sin x+2\cos x+c$
2. $\int x^2\sin xdx$

   Solution 👉 Click Here

   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 Derivative 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
   $\int x^2\sin xdx=-x^2\cos x+2x\sin x+2\cos x$
3. $\int x^2{\sin }^{-1}xdx$

   Solution 👉 Click Here

   Solution
   By ILATE rule, we let $u={\sin }^{-1}x,v=x^2$, in which $u^{\prime }=\frac{1}{\sqrt{1-x^2}},\int v=\frac{x^3}{3}$
   Hence
   $I=\int uvdx$
   or $I=u\int v-\int (u^{\prime }\int v)$
   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 $I=\frac{1}{3}{x}^3{\sin }^{-1}x+\frac{1}{6}\int (\frac{1}{\sqrt{s}}-\sqrt{s})ds$
   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 $\sqrt{x}=u$

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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	$\int x{e}^{x^2}dx$	$\frac{1}{2}{e}^{x^2}+c$
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	$\int x{e}^{x}dx$	$(x-1)e^x+c$
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	$\int x\sin x\cos xdx$	$\frac{x}{4}-\frac{x{\cos }^{2}x}{2}+\frac{\cos x\sin x}{4}+c$
12	$\int {\sin }^{2}xdx$	$\frac{x}{2}-\frac{\sin (2x)}{4}+c$
13	$\int x{\sin }^{2}xdx$	$\frac{x^2}{4}-\frac{{\cos }^{2}x}{4}-\frac{x\sin x\cos x}{2}+c$