Sections:
- Constants & single algebraic terms
- $ f'(x) [f(x)]^n$
- Exponential terms
- Trigonometric terms
- Trigonometric terms & Double angle formula
- Trigonometric terms & Pythagoras identities
- Fractions
- Integration by parts
- Integration by substitution
(1) Constants & single algebraic terms
Techniques:
If $a$ is constant,
$ \int a \phantom{.} dx = $
$ ax + C $
If $a$ is constant and $n \ne -1$,
$ \int ax^{n} \phantom{.} dx = $
$ {a \over n + 1} x^{n+ 1} + C $
If $a$ is constant,
$ \int ax^{-1} \phantom{.} dx = \int {a \over x} \phantom{.} dx = $
$ a \ln |x| + C $
(2) $ f'(x) [f(x)]^n$
Technique 1 ($n \ne -1$):
If $n \ne -1$,
$ \int f'(x) [f(x)]^n \phantom{.} dx = $
$ { [f(x)]^{n + 1} \over (n + 1) } + C $
Example
Integrate $ {x + 1 \over (x^2 + 2x - 5)^2 }$ with respect to $x$.
Answer: $ - {1 \over 2(x^2 + 2x - 5)} + C $
Solutions
\begin{align*}
[f(x) = x^2 + 2x - 5, & f'(x) = 2x + 2] \\
\\
\int {x + 1 \over (x^2 + 2x - 5)^2} \times {2 \over 2} \phantom{.} dx & = {1 \over 2} \int {2x + 2 \over (x^2 + 2x - 5)^2} \phantom{.} dx \\
& = {1 \over 2} \int (2x + 2) (x^2 + 2x - 5)^{-2} \phantom{.} dx \\
& = {1 \over 2} \left[ (x^2 + 2x - 5)^{-1} \over -1 \right] \\
& = -{1 \over 2(x^2 + 2x - 5)} + C
\end{align*}
Technique 2 ($n = - 1$):
$ \int f'(x) [f(x)]^{-1} \phantom{.} dx = \int {f'(x) \over f(x)} \phantom{.} dx = $
$ \ln |f(x)| + C $
Example
Integrate $ {4x + 4 \over x^2 + 2x - 5} $ with respect to $x$.
Answer: $ 2 \ln |x^2 + 2x - 5|+ C $
Solutions
\begin{align*}
[f(x) = x^2 + 2x - 5, & f'(x) = 2x + 2] \\
\\
\int {4x + 4 \over x^2 + 2x - 5} \phantom{.} dx & = {2 \over 1} \int {2x + 2 \over x^2 + 2x - 5} \phantom{.} dx \\
& = 2 \ln |x^2 + 2x - 5| + C
\end{align*}
(3) Exponential terms
Technique:
$ \int f'(x) . e^{f(x)} \phantom{.} dx = $
$ e^{f(x)} + C $
Example 1
Integrate $ \sin x e^{\cos x} $ with respect to $x$.
Answer: $ - e^{\cos x} + C $
Solutions
\begin{align*}
[f(x) = \cos x, & f'(x) =- \sin x] \\
\\
\int \sin x \phantom{.} e^{\cos x} \phantom{.} dx & = - \int - \sin x \phantom{.} e^{\cos x} \phantom{.} dx \\
& = -e^{\cos x} + C
\end{align*}
Example 2
Integrate $ {e^{3x} \over 1 - e^{3x}} $ with respect to $x$.
Hint: Need to use integration formula from (2).
Answer: $ -{1 \over 3} \ln | 1 - e^{3x} | + C $
Solutions
\begin{align*}
[f(x) = 1 - e^{3x}, & f'(x) =0 - 3e^{3x} = - 3e^{3x}] \\
\\
\int {e^{3x} \over 1 - e^{3x} } \times {-3 \over -3} \phantom{.} dx & = -{1 \over 3} \int { -3e^{3x} \over 1 - e^{3x} } \phantom{.} dx \phantom{00000} \left[ \int {f'(x) \over f(x)} \phantom{.} dx \right] \\
& = -{1 \over 3} \ln |1 - e^{3x}| + C
\end{align*}
(4) Trigonometric terms
Techniques:
$ f'(x) . \sin [f(x)] \phantom{.} dx = $
$ -\cos [f(x)] + C $
$ f'(x) . \cos [f(x)] \phantom{.} dx = $
$ \sin [f(x)] + C $
$ f'(x) . \sec^2 [f(x)] \phantom{.} dx = $
$ \tan [f(x)] + C $
$ f'(x) . \text{cosec}^2 [f(x)] \phantom{.} dx = $
$ - \cot [f(x)] + C $
$ f'(x) . \text{cosec}^2 [f(x)] \phantom{.} dx = $
$ - \cot [f(x)] + C $
$ f'(x) . \sec [f(x)] \tan [f(x)] \phantom{.} dx = $
$ \sec [f(x)] + C $
$ f'(x) . \text{cosec } [f(x)] \cot [f(x)] \phantom{.} dx = $
$ - \text{cosec } [f(x)] + C $
From MF27, if $ \int \tan [f(x)] \phantom{.} dx = \ln | \sec x | + C $, then
$ \int f'(x) . \tan [f(x)] \phantom{.} dx = $
$ \ln | \sec [f(x)] | + C $
From MF27, if $ \int \cot [f(x)] \phantom{.} dx = \ln | \sin x | + C $, then
$ \int f'(x) . \cot [f(x)] \phantom{.} dx = $
$ \ln | \sin [f(x)] | + C $
From MF27, if $ \int \text{cosec } [f(x)] \phantom{.} dx = - \ln | \text{cosec } x + \cot x | + C $, then
$ \int f'(x) . \cot [f(x)] \phantom{.} dx = $
$ - \ln | \text{cosec } [f(x)] + \cot [f(x)] | + C $
From MF27, if $ \int \sec x \phantom{.} dx = \ln | \sec x + \tan x | + C $, then
$ \int f'(x) . \sec [f(x)] \phantom{.} dx = $
$ \ln | \sec [f(x)] + \tan [f(x)] | + C $
Example 1
Integrate $ \cos 3x \sin^3 3x $ with respect to $x$.
Hint: Need to use integration formula from (2).
Answer: $ {1 \over 12} \sin^4 3x + C $
Solutions
\begin{align*}
& [ f(x) = \sin 3x, f'(x) = 3 \cos 3x] \\
\\
\int \cos 3x \sin^3 3x \phantom{.} dx
& = {1 \over 3} \int 3 \cos 3x (\sin 3x)^3 \phantom{.} dx
\phantom{000000} \left[ \int f'(x) [f(x)]^n \phantom{.} dx \right] \\
& = {1 \over 3} \left[ (\sin 3x)^4 \over 4 \right] \\
& = {1 \over 12} \sin^4 3x + C
\end{align*}
Example 2
Integrate $ \cot {1 \over 2}x \sec^2 {1 \over 2}x $ with respect to $x$.
Hint: Need to use integration formula from (2).
Answer: $ 2 \ln | \tan {1 \over 2} x | + C $
Solutions
\begin{align*}
\int \cot {1 \over 2}x \sec^2 {1 \over 2}x \phantom{.} dx
& = \int {1 \over \tan {1 \over 2}x} . \sec^2 {1 \over 2}x \phantom{.} dx \\
& = \int { \sec^2 {1 \over 2}x \over \tan {1 \over 2}x } \phantom{.} dx \\
& = 2 \int { {1 \over 2} \sec^2 {1 \over 2}x \over \tan {1 \over 2}x } \phantom{.} dx
\phantom{0000} [ f(x) = \tan {1 \over 2}x, f'(x) = {1 \over 2} \sec^2 {1 \over 2}x ] \\
& = 2 \ln \left| \tan {1 \over 2} x \right| + C
\end{align*}
(5) Trigonometric terms & Double angle formula
Double angle formulas in MF27:
$$ \sin 2A = 2 \sin A \cos A $$
$$ \cos 2A = 2 \cos^2 A - 1 = 1 - 2 \sin^2 A $$
Example 1
Integrate $ \sin 3x \cos 3x $ with respect to $x$.
Answer: $ -{1 \over 12} \cos 6x + C $
Solutions
\begin{align*}
\int \sin 3x \cos 3x \phantom{.} dx & = {1 \over 2} \int 2 \sin 3x \cos 3x \phantom{.} dx \\
& = {1 \over 2} \int \sin 6x \phantom{.} dx \phantom{00000000} [\sin 2A = 2 \sin A \cos A] \\
& = {1 \over 12} \int 6 \sin 6x \phantom{.} dx \phantom{00000} \left[ \int f'(x) \sin [f(x)] \phantom{.} dx \right]\\
& = {1 \over 12} (-\cos 6x) \\
& = -{1 \over 12} \cos 6x + C
\end{align*}
Example 2
Integrate $ \sin^2 2x $ with respect to $x$.
Answer: $ {1 \over 2}x - {1 \over 8} \sin 4x + C $
Solutions
\begin{align*}
\text{Since } \cos 2A & = 1 - 2 \sin^2 A, \\
2 \sin^2 A & = 1 - \cos 2A \\
\sin^2 A & = {1 \over 2} - {1 \over 2} \cos 2A \\
\\
\int \sin^2 2x \phantom{.} dx & = \int \left( {1 \over 2} - {1 \over 2} \cos 4x \right) \phantom{.} dx \\
& = \int {1 \over 2} \phantom{.} dx - {1 \over 2} \int \cos 4x \phantom{.} dx \\
& = {1 \over 2}x - {1 \over 8} \int 4 \cos 4x \phantom{.} dx \\
& = {1 \over 2}x - {1 \over 8} \sin 4x + C
\end{align*}
Example 3
Integrate $ 2 \cos^2 \left(x \over 3\right) $ with respect to $x$.
Answer: $ {3 \over 2} \sin \left({2 \over 3}x\right) + x + C $
Solutions
\begin{align*}
\text{Since } \cos 2A & = 2 \cos^2 A - 1, \\
- 2 \cos^2 A & = - \cos 2A - 1 \\
2 \cos^2 A & = \cos 2A + 1 \\
\\
\int 2 \cos^2 \left(x \over 3\right) \phantom{.} dx
& = \int \cos \left({2 \over 3}x\right) + 1 \phantom{.} dx \\
& = \int \cos \left( {2 \over 3}x \right) \phantom{.} dx + \int 1 \phantom{.} dx \\
& = {3 \over 2} \int {2 \over 3} \cos \left( {2 \over 3}x \right) \phantom{.} dx + x \\
& = {3 \over 2} \sin \left( {2 \over 3}x \right) + x + C
\end{align*}
(6) Trigonometric terms & Pythagoras identities
Pythagoras identities (need to know):
$$ \sec^2 A = 1 + \tan^2 A $$
$$ \text{cosec}^2 A = 1 + \cot^2 A $$
Example 1
Integrate $ 2 \tan^2 {x \over 3} $ with respect to $x$.
Answer: $ 6 \tan {x \over 3} - 2x + C $
Solutions
\begin{align*}
\text{Since } \sec^2 A & = 1 + \tan^2 A, \\
\tan^2 A & = \sec^2 A - 1 \\
\\
\int 2 \tan^2 {x \over 3} \phantom{.} dx & = \int 2 \left( \sec^2 {x \over 3} - 1 \right) \phantom{.} dx \\
& = \int 2 \sec^2 {x \over 3} \phantom{.} dx - \int 2 \phantom{.} dx \\
& = 6 \int {1 \over 3} \sec^2 {x \over 3} \phantom{.} dx - 2x \\
& = 6 \tan {x \over 3} - 2x + C
\end{align*}
Example 2
Integrate $ \cot^2 (\pi x ) $ with respect to $x$.
Answer: $ -{1 \over \pi} \cos (\pi x) - x + C $
Solutions
\begin{align*}
\text{Since } \text{cosec}^2 A & = 1 + \cot^2 A, \\
\cot^2 A & = \text{cosec}^2 A - 1 \\
\\
\int \cot^2 (\pi x) \phantom{.} dx & = \int \text{cosec}^2 (\pi x) - 1 \phantom{.} dx \\
& = \int \text{cosec}^2 (\pi x) \phantom{.} dx - \int 1 \phantom{.} dx \\
& = {1 \over \pi} \int \pi \text{ cosec}^2 (\pi x) \phantom{.} dx - x \\
& = {1 \over \pi} [ -\cot (\pi x) ] - x \\
& = -{1 \over \pi} \cot (\pi x) - x + C
\end{align*}
(7) Fractions
Techniques in MF27:
$$ \int {1 \over x^2 + a^2} \phantom{.} dx = {1 \over a} \tan^{-1} \left(x \over a\right) $$
$$ \int {1 \over \sqrt{a^2 - x^2} } \phantom{.} dx = \sin^{-1} \left(x \over a\right) $$
$$ \int {1 \over x^2 - a^2} \phantom{.} dx = {1 \over 2a} \ln \left| {x - a \over x + a} \right| $$
$$ \int {1 \over a^2 - x^2} \phantom{.} dx = {1 \over 2a} \ln \left| {a + x \over a - x} \right| $$
Example 1
Integrate $ {1 \over \sqrt{3 - 25x^2} } $ with respect to $x$.
Answer: $ {1 \over 5} \sin^{-1} \left(5x \over \sqrt{3}\right) + C $
Solutions
\begin{align*}
\int {1 \over \sqrt{3 - 25x^2} } \phantom{.} dx & = \int { 1 \over \sqrt{25} \sqrt{ {3 \over 25} - x^2 } } \phantom{.} dx
\phantom{000000} [\text{Ensure coefficient of } x^2 \text{ is 1}] \\
& = {1 \over 5} \int { 1 \over \sqrt{ \left(\sqrt{3} \over 5\right)^2 - x^2 } } \phantom{.} dx \\
& = {1 \over 5} \sin^{-1} \left( x \over {\sqrt{3} \over 5} \right) \\
& = {1 \over 5} \sin^{-1} \left(5x \over \sqrt{3} \right) + C
\end{align*}
Example 2
Integrate $ {1 \over x^2 - 3x + 2 } $ with respect to $t$.
Answer: $ \ln \left| {x - 2 \over x - 1} \right|+ C $
Solutions (Method 1)
\begin{align*}
x^2 - 3x + 2 & = x^2 - 3x + \left(3 \over 2\right)^2 - \left(3 \over 2\right)^2 + 2 \\
& = \left(x - {3 \over 2}\right)^2 - {1 \over 4} \phantom{00000000000000} [\text{Complete the square}] \\
\\
\int {1 \over x^2 - 3x + 2} \phantom{.} dx & = \int {1 \over \left(x - {3 \over 2}\right)^2 - {1 \over 4}} \phantom{.} dx \\
& = \int {1 \over \left(x - {3 \over 2}\right)^2 - \left(1 \over 2\right)^2} \phantom{.} dx \\
& = {1 \over 2\left(1 \over 2\right)} \ln \left| x - {3 \over 2} - {1 \over 2} \over x - {3 \over 2} + {1 \over 2} \right| \\
& = \ln \left| {x - 2 \over x - 1} \right| + C
\end{align*}
Solutions (Method 2)
\begin{align*}
{1 \over x^2 - 3x + 2} & = {1 \over (x - 1)(x - 2)} \\
& = {A \over x - 1} + {B \over x - 2} \\
& = {A (x - 2) + B (x - 1) \over (x - 1)(x - 2)} \\
\\
1 & = A(x - 2) + B(x - 1) \\
\\
\text{Let } & x = 2, \\
1 & = B \\
\\
\text{Let } & x = 1, \\
1 & = -A \\
-1 & = A \\
\\
{1 \over x^2 - 3x + 2} & = {-1 \over x - 1} + {1 \over x - 2} \\
\\
\int {1 \over x^2 - 3x + 2} \phantom{.} dx & = \int \left( {-1 \over x - 1} + {1 \over x - 2} \right) \phantom{.} dx \\
& = - \int {1 \over x - 1} \phantom{.} dx + \int {1 \over x - 2} \phantom{.} dx \\
& = - \ln |x - 1| + \ln |x - 2| \\
& = \ln \left| {x - 2 \over x - 1} \right| + C
\end{align*}
Proper fractions in the form ${f(x) \over g(x)}$:
Example
Integrate $ {2x \over \sqrt{5 - 4x - x^2} } $ with respect to $x$.
Answer: $ -2\sqrt{5 - 4x -x^2} - 4 \sin^{-1} \left(x + 2 \over 3\right) + C $
Solutions
\begin{align*}
f(x) & = 5 - 4x - x^2, f'(x) = -4 - 2x \\
\\
\int { 2x \over \sqrt{5 - 4x - x^2} } \phantom{.} dx & = - \int {-2x \over \sqrt{5 - 4x - x^2} } \phantom{.} dx \\
& = - \int \left( {- 2x - 4 \over \sqrt{5 - 4x - x^2} } + {4 \over \sqrt{5 - 4x - x^2} } \right) \phantom{.} dx \\
& = - \int {-2x - 4 \over \sqrt{5 - 4x - x^2} } \phantom{.} dx - \int {4 \over \sqrt{5 - 4x - x^2} } \phantom{.} dx \\
& = - \int (-2x - 4) (5 - 4x - x^2)^{-{1 \over 2}} \phantom{.} dx - 4 \int {1 \over \sqrt{5 - 4x - x^2} } \phantom{.} dx \\
\\
\int (-2x - 4) (5 - 4x - x^2)^{-{1 \over 2}} \phantom{.} dx & = {(5 - 4x - x^2)^{1 \over 2} \over {1 \over 2}} \\
& = 2 \sqrt{ 5 - 4x - x^2 } \\
\\
5 - 4x - x^2 & = -x^2 - 4x + 5 \\
& = -(x^2 + 4x) + 5 \\
& = - [ (x + 2)^2 - (2)^2 ] + 5 \\
& = -(x + 2)^2 + 4 + 5 \\
& = -(x + 2)^2 + 9 \\
& = 9 - (x + 2)^2 \\
& = 3^2 - (x + 2)^2 \\
\\
\int {1 \over \sqrt{5 - 4x - x^2} } \phantom{.} dx & = \int {1 \over \sqrt{3^2 - (x + 2)^2}} \phantom{.} dx \\
& = \sin^{-1} \left(x + 2 \over 3\right) \\
\\
\therefore \text{Required answer} & = - 2 \sqrt{ 5 - 4x - x^2 } - 4 \sin^{-1} \left(x + 2 \over 3\right) + C
\end{align*}
Improper fractions in the form ${f(x) \over g(x)}$:
Example
Integrate $ {2x^2 \over x^2 + 3x + 5} $ with respect to $x$.
Answer: $ 2x - 3 \ln (x^2 + 3x + 5) - {2 \over \sqrt{11} } \tan^{-1} \left(2x + 3 \over \sqrt{11}\right)+ C $
Solutions
\begin{align*}
f(x) & = x^2 + 3x + 5, f'(x) = 2x + 3 \\
\\
\int {2x^2 \over x^2 + 3x + 5} \phantom{.} dx & = \int 2 + {-6x - 10 \over x^2 + 3x + 5} \phantom{.} dx \\
& = \int 2 \phantom{.} dx - \int {6x + 10 \over x^2 + 3x + 5} \phantom{.} dx \\
& = 2x - \int \left( {6x + 9 \over x^2 + 3x + 5} + {1 \over x^2 + 3x + 5} \right) \phantom{.} dx \\
& = 2x - \int {6x + 9 \over x^2 + 3x + 5} \phantom{.} dx - \int {1 \over x^2 + 3x + 5} \phantom{.} dx \\
& = 2x - 3\int {2x + 3 \over x^2 + 3x + 5} \phantom{.} dx - \int {1 \over\left(x + {3 \over 2}\right)^2 - \left(3 \over 2\right)^2 + 5} \phantom{.} dx \\
& = 2x - 3\int {2x + 3 \over x^2 + 3x + 5} \phantom{.} dx - \int {1 \over \left(x + {3 \over 2}\right)^2 + {11 \over 4} } \phantom{.} dx \\
& = 2x - 3\int {2x + 3 \over x^2 + 3x + 5} \phantom{.} dx - \int {1 \over \left(x + {3 \over 2}\right)^2 + \left(\sqrt{11} \over 2\right)^2 } \phantom{.} dx \\
& = 2x - 3 \ln (x^2 + 3x + 5) - {1 \over {\sqrt{11} \over 2}} \tan^{-1} \left(x + {3 \over 2} \over {\sqrt{11} \over 2} \right) \\
& = 2x - 3 \ln (x^2 + 3x + 5) - {2 \over \sqrt{11}} \tan^{-1} \left(2x + 3 \over \sqrt{11}\right) + C
\end{align*}
(8) Integration by parts
Formula:
$ \int u {dv \over dx} \phantom{.} dx = $ $ uv - \int v {du \over dx} \phantom{.} dx $
Priority for selecting $u$ (highest to lowest):
- Logarithms
- Inverse trigonometry
- Algebra
- Trigonometry
- Exponential
Example 1
Integrate $ \tan^{-1} x $ with respect to $x$.
Answer: $ x \tan^{-1} x - {1 \over 2} \ln (1 + x^2) + c $
Solutions
\begin{align*}
\int \tan^{-1} x \phantom{.} dx & = \int (\tan^{-1} x) (1) \phantom{.} dx \phantom{0000000} [\text{Considered algebra}] \\
\\
u & = \tan^{-1} x \phantom{000000} {dv \over dx} = 1 \\
{du \over dx} & = {1 \over 1 + x^2} \phantom{00000000} v = x \\
\\
\int \tan^{-1} x \phantom{.} dx & = \int (\tan^{-1} x) (1) \phantom{.} dx \\
& = (\tan^{-1} x)(x) - \int (x)\left(1 \over 1 + x^2\right) \phantom{.} dx \\
& = x \tan^{-1} x - \int {x \over 1 + x^2} \phantom{.} dx \\
& = x \tan^{-1} x - {1 \over 2} \int {2x \over 1 + x^2} \phantom{.} dx
\phantom{000000} \left[ \int {f'(x) \over f(x)} \phantom{.} dx \right] \\
& = x \tan^{-1} x - {1 \over 2} \ln (1 + x^2) + c
\end{align*}
Example 2: By parts more than once
Integrate $ x^2 e^{3x} $ with respect to $x$.
Answer: $ {1 \over 3} x^2 e^{3x} - {2 \over 9} x e^{3x} + {2 \over 27} e^{3x} + c $
Solutions
\begin{align*}
u & = x^2 \phantom{000000} {dv \over dx} = e^{3x} \\
{du \over dx} & = 2x \phantom{00000000} v = {1 \over 3}e^{3x} \\
\\
\int x^2 e^{3x} \phantom{.} dx
& = x^2 \left({1 \over 3}e^{3x}\right) - \int \left({1 \over 3}e^{3x}\right) (2x) \phantom{.} dx \\
& = {1 \over 3} x^2 e^{3x} - {2 \over 3} \underbrace{\int x e^{3x} \phantom{.} dx}_{\text{By parts again}} \\
\\
u & = x \phantom{000000} {du \over dx} = 1 \\
{dv \over dx} & = e^{3x} \phantom{000000.} v = {1 \over 3}e^{3x} \\
\\ \\
\int x e^{3x} \phantom{.} dx & = (x)\left({1 \over 3}e^{3x}\right)
- \int \left({1 \over 3}e^{3x}\right) (1) \phantom{.} dx \\
& = {1 \over 3} x e^{3x} - {1 \over 3} \int e^{3x} \phantom{.} dx \\
& = {1 \over 3} x e^{3x} - {1 \over 3} \left( {1 \over 3} e^{3x} \right) \\
& = {1 \over 3} x e^{3x} - {1 \over 9} e^{3x} \\
\\ \\
\therefore {1 \over 3} x^2 e^{3x} - {2 \over 3} \int x e^{3x} \phantom{.} dx
& = {1 \over 3} x^2 e^{3x} - {2 \over 3} \left( {1 \over 3} x e^{3x} - {1 \over 9} e^{3x} \right) \\
& = {1 \over 3} x^2 e^{3x} - {2 \over 9} x e^{3x} + {2 \over 27} e^{3x} + c
\end{align*}
Example 3: Change subject of equation
Integrate $ e^{x} \cos x $ with respect to $x$.
Answer: $ {1 \over 2} e^x \cos x + {1 \over 2} e^x \sin x + c $
Solutions
\begin{align*}
u & = \cos x \phantom{0000000000.} v = e^{x} \\
{du \over dx} & = - \sin x \phantom{0000000} {dv \over dx} = e^{x} \\
\\
\int e^x \cos x \phantom{.} dx & = (\cos x)(e^x) - \int (e^x)(- \sin x) \phantom{.} dx \\
& = e^x \cos x + \underbrace{\int e^x \sin x \phantom{.} dx}_\text{By parts again} \\
\\
u & = \sin x \phantom{000000} {du \over dx} = \cos x \\
{dv \over dx} & = e^{x} \phantom{0000000.} v = e^{x} \\
\\
\therefore \int e^x \cos x \phantom{.} dx
& = e^x \cos x + (\sin x)(e^x) - \int (e^x)(\cos x) \phantom{.} dx \\
\underbrace{\int e^x \cos x \phantom{.} dx}_\text{Same}
& = e^x \cos x + e^x \sin x - \underbrace{\int e^x \cos x \phantom{.} dx}_\text{Same} \\
2 \int e^x \cos x \phantom{.} dx & = e^x \cos x + e^x \sin x \\
\int e^x \cos x \phantom{.} dx & = {1 \over 2} e^x \cos x + {1 \over 2} e^x \sin x + c
\end{align*}
(9) Integration by substitution
Indefinite integral:
Steps:
- Change the expression from $x$ to $u$
- Change $dx$ to $du$
- Integrate and express answer in terms of $x$
Example 1
Using the the substitution $u = e^x$, find $ \int {e^x \over \sqrt{1 - e^{2x}} } \phantom{.} dx $.
Answer: $ \sin^{-1} (e^x) + c $
Solutions
\begin{align*}
u & = e^x \\
\\
{du \over dx} & = e^x \\
{du \over e^x} & = dx \\
{du \over u} & = dx \\
\\
\therefore \int {e^x \over \sqrt{1 - e^{2x}} } \phantom{.} dx
& = \int {u \over \sqrt{1 - u^2} } \phantom{.} {du \over u}
\phantom{000000} [\text{Change from } x \text{ to } u \text{ and } dx \text{ to } du] \\
& = \int {1 \over \sqrt{1 - u^2} } \phantom{.} du \\
& = \sin^{-1} u + c \\
& = \sin^{-1} (e^x) + c \phantom{000000000} [\text{Change back to in terms of } x]
\end{align*}
Example 2: Trigonometry
Using the substitution $x = 4 \sin \theta$, find $ \int \sqrt{16 - x^2} \phantom{.} dx $.
Answer: $ {1 \over 2} x \sqrt{16 - x^2} + 8 \sin^{-1} \left(x \over 4\right) + c $
Solutions
\begin{align*}
x & = 4 \sin \theta \\
{dx \over d \theta} & = 4 \cos \theta \\
dx & = 4 \cos \theta \phantom{.} d \theta \\
\\
\int \sqrt{16 - x^2} \phantom{.} dx
& = \int \sqrt{ 16 - (4 \sin \theta)^2 } \phantom{.} 4 \cos \theta \phantom{.} d \theta \\
& = \int \sqrt{ 16 - 16\sin^2 \theta} \phantom{.} 4 \cos \theta \phantom{.} d \theta \\
& = \int 4 \sqrt{1 - \sin^2 \theta } \phantom{.} 4 \cos \theta \phantom{.} d \theta \\
& = \int 16 \sqrt{\cos^2 \theta} \phantom{.} \cos \theta \phantom{.} d \theta
\phantom{000000} [\text{Identity: } \sin^2 A + \cos^2 A = 1] \\
& = \int 16 \cos^2 \theta \phantom{.} d \theta \\
& = \int 8 (2 \cos^2 \theta) \phantom{.} d \theta \\
& = \int 8 (\cos 2\theta + 1) \phantom{.} d \theta
\phantom{000000000} [\text{Double angle formula}] \\
& = \int 8 \cos 2 \theta + 8 \phantom{.} d \theta \\
& = \int 4 (2 \cos 2\theta) + 8 \phantom{.} d \theta \\
& = 4 \sin 2 \theta + 8 \theta \\
& = 4 (2 \sin \theta \cos \theta) + 8 \theta
\phantom{000000000} [\text{Double angle formula}] \\
& = 8 \sin \theta \cos \theta + 8 \theta
\end{align*}
\begin{align*}
x & = 4 \sin \theta \\
{x \over 4} & = \sin \theta \\
{x \over 4} & = {Opp \over Hyp} \\
\\
\therefore \int \sqrt{16 - x^2} \phantom{.} dx
& = 8 \left(x \over 4\right)\left(\sqrt{16 - x^2} \over 4\right) + 8\sin^{-1} \left(x \over 4\right) \\
& = {8x \over 16} \sqrt{16 - x^2} + 8 \sin^{-1} \left(x \over 4\right) \\
& = {1 \over 2} x \sqrt{16 - x^2} + 8 \sin^{-1} \left(x \over 4\right) + c
\end{align*}
Definite integral:
Steps:
- Change limits of integration ($x$ to $\theta$)
- Change the expression from $x$ to $\theta$
- Change $dx$ to $d\theta$
- Integrate
Example
Using the substitution $x = \tan \theta$, find $ \int_{0}^1 {1 - x^2 \over (1 + x^2)^2} \phantom{.} dx $.
Answer: $ {1 \over 2} x \sqrt{16 - x^2} + 8 \sin^{-1} \left(x \over 4\right) + c $
Solutions
\begin{align*}
x & = \tan \theta \\
\tan^{-1} (x) & = \theta \\
\\
\text{When } & x = 1, \\
\theta & = \tan^{-1} (1) = {\pi \over 4} \\
\\
\text{When } & x = 0, \\
\theta & = \tan^{-1} (0) = 0 \\
\\
{dx \over d\theta} & = \sec^2 \theta \\
dx & = \sec^2 \theta \phantom{.} d \theta \\
\\
\int_0^1 {1 - x^2 \over (1 + x^2)^2} \phantom{.} dx & =
\int_0^{\pi \over 4} {1 - \tan^2 \theta \over (1 + \tan^2 \theta)^2} \sec^2 \theta \phantom{.} d \theta \\
& = \int_0^{\pi \over 4} {1 - \tan^2 \theta \over (\sec^2 \theta)^2} \sec^2 \theta \phantom{.} d \theta
\phantom{000000} [\text{Identity: } \tan^2 A + 1 = \sec^2 A] \\
& = \int_0^{\pi \over 4} {1 - \tan^2 \theta \over \sec^2 \theta} \phantom{.} d \theta \\
& = \int_0^{\pi \over 4} {1 \over \sec^2 \theta} - {\tan^2 \theta \over \sec^2 \theta} \phantom{.} d\theta \\
& = \int_0^{\pi \over 4} \cos^2 \theta - {\sin^2 \theta \over \cos^2 \theta} \times \cos^2 \theta
\phantom{.} d \theta \\
& = \int_0^{\pi \over 4} \cos^2 \theta - \sin^2 \theta \phantom{.} d\theta \\
& = \int_0^{\pi \over 4} \cos 2 \theta \phantom{.} d\theta
\phantom{000000} [\text{Double angle formula}] \\
& = {1 \over 2} \int_0^{\pi \over 4} 2 \cos 2 \theta \phantom{.} d\theta \\
& = {1 \over 2} \left[ \sin 2 \theta \right]_0^{\pi \over 4} \\
& = {1 \over 2} \left\{ \sin \left[2 \left(\pi \over 4\right)\right] - \sin [2(0)] \right\} \\
& = {1 \over 2} (1 - 0) \\
& = {1 \over 2}
\end{align*}