Substitution Integrals

About Substitution Integral

Sometimes solving integrals of the form $\int {f\left( x \right)} dx$ requires special techniques. One of the special techniques is to use the integral substitution formula. There are two kinds of substitution integral formulas that are commonly used, namely:

1. Modifiable integration in the form $\int {f\left( u \right)} du$
2. Integration containing the forms $\sqrt {{a^2} - {x^2}} $, $\sqrt {{a^2} + {x^2}} $, and $\sqrt { {x^2} - {a^2}} $

A mutable integration of the form $\int {f\left( u \right)} du$

Suppose using the substitution $u = g\left( x \right)$, where $g$ is a function that has a derivative, so $\int {f\left( {g\left( x \right )} \right)} g'\left( x \right)dx$ can be changed to $\int {f\left( u \right)} du$. If ${f\left( u \right)}$ is anti-differential from ${f\left( x \right)}$ then:

$\boxed{\int {f\left( {g\left( x \right)} \right)} g'\left( x \right)dx = \int {f\left( u \right)} du = F\left( u \right) + C = F\left( {g\left( x \right)} \right) + C}$

The integration calculation technique using the substitution integral formula requires two steps as follows:

1. Choose a function $u = g\left( x \right)$ so that $\int {f\left( {g\left( x \right)} \right)} g'\left( x \right)dx $ can be changed to $\int {f\left( u \right)} du$.
2. Find the general integral function ${f\left( u \right)}$ which is $F'\left( {du} \right) = f\left( u \right)$.

To determine the general integral function $f\left( u \right)$ which is $F'\left( {du} \right) = f\left( u \right)$, can be obtained by developing the formulas Indeterminate integrals of algebraic and trigonometric functions. The development formulas can be summarized as follows:

1. Integrating Algebraic Functions
$\boxed{\int {{u^n}} du = \frac{1}{{n + 1}}{u^{n + 1}} + C},n \in Rasional,n \ne 1$
2. Integrating Trigonometric Functions
• $\int {\sin udu = - \cos u + C} $
• $\int {\cos udu = \sin u + C} $
• $\int {{{\sec }^2}udu = \tan u + C} $
• $\int {\cos e{c^2}udu = - \cot u + C} $
• $\int {\tan u.\sec udu = \sec u + C} $
• $\int {\cot u.\cos ecudu = - \cos ecu + C} $

Examples

Find the following indefinite integrals

1. $\int {{{\left( {4x + 5} \right)}^6}} dx$
2. ${\int {10x\left( {8{x^2} - 1} \right)} ^4}dx$
3. $\int {\sqrt {\sin x} } \cos xdx$
4. $\int {\frac{{\sin x}}{{{{\cos }^2}x}}} dx$

Answers

1. For example $u = 4x + 5$ then $du = 4dx$ or $dx = \frac{1}{4}du$
$\begin{array}{l} \int {{{\left( {4x + 5} \right)}^6}} dx = \int {{u^6}} .\frac{1}{4}du\\ = \frac{1}{4}\int {{u^6}} du\\ = \frac{1}{4}\left( {\frac{1}{7}{u^7}} \right) + C\\ = \frac{1}{{28}}{u^7} + C\\ = \frac{1}{{28}}{\left( {4x + 5} \right)^7} + C \end{array}$

So, $\int {{{\left( {4x + 5} \right)}^6}} dx = \frac{1}{{28}}{\left( {4x + 5} \right )^7} + C$.

2. For example $u = 8{x^2} - 1$ then $du = 16xdx$ or $dx = \frac{{du}}{{16x}}$
$\begin{array}{l} {\int {10x\left( {8{x^2} - 1} \right)} ^4}dx = {\int {10x\left( u \right)} ^4}.\frac{{du} }{{16x}}\\ = \int {\frac{{10}}{{16}}} {u^4}du = \int {\frac{5}{8}} {u^4}du\\ = \frac{5}{8}\int {{u^4}du} \\ = \frac{5}{8}\left( {\frac{1}{5}{u^5}} \right) + C\\ = \frac{1}{8}{u^5} + C\\ = \frac{1}{8}{\left( {8{x^2} - 1} \right)^5} + C \end{array}$

So, ${\int {10x\left( {8{x^2} - 1} \right)} ^4}dx = \frac{1}{8}{\left( {8{x^ 2} - 1} \right)^5} + C$.

3. For example $u = \sin x$ then $du = \cos xdx$ or $dx = \frac{{du}}{{\cos x}}$
$\begin{array}{l} \int {\sqrt {\sin x} } \cos xdx = {\int {\left( {\sin x} \right)} ^{\frac{1}{2}}}\cos xdx\\ = \int {{u^{\frac{1}{2}}}} \cos x.\frac{{du}}{{\cos x}}\\ = \int {{u^{\frac{1}{2}}}} du\\ = \frac{1}{{\frac{1}{2} + 1}}{u^{\frac{1}{2} + 1}} + C\\ = \frac{2}{3}{u^{\frac{3}{2}}} + C\\ = \frac{2}{3}u\sqrt u + C\\ = \frac{2}{3}\sin x\sqrt {\sin x} + C \end{array}$

So, $\int {\sqrt {\sin x} } \cos xdx = \frac{2}{3}\sin x\sqrt {\sin x} + C$.

4. For example $u = \cos x$ then $du = - \sin xdx$ or $dx = - \frac{{du}}{{\sin x}}$
$\begin{array}{l} \int {\frac{{\sin x}}{{{{\cos }^2}x}}} dx = \int {\sin x.{{\cos }^{ - 2}}xdx} \\ = \int {\sin x.{u^{ - 2}}} . - \frac{{du}}{{\sin x}}\\ = - \int {{u^{ - 2}}du} \\ = - \left( {\frac{1}{{ - 2 + 1}}{u^{ - 2 + 1}}} \right) + C\\ = - \left( { - {u^{ - 1}}} \right) + C\\ = {u^{ - 1}} + C\\ = \frac{1}{u} + C\\ = \frac{1}{{\cos x}} + C \end{array}$

So $\int {\frac{{\sin x}}{{{{\cos }^2}x}}} dx = \frac{1}{{\cos x}} + C$.

Integration containing the forms $\sqrt {{a^2} - {x^2}} $, $\sqrt {{a^2} + {x^2}} $, and $\sqrt { {x^2} - {a^2}} $

Integration containing the forms $\sqrt {{a^2} - {x^2}} $, $\sqrt {{a^2} + {x^2}} $, and $\sqrt { {x^2} - {a^2}} $ can be solved using integral substitution as shown in the following table:

Integral function	Substitution	Substitution Result
$\sqrt {{a^2} - {x^2}} $	$x = a\sin \theta $	$a\sqrt {1 - {{\sin }^2}\theta } = a\cos \theta $
$\sqrt {{a^2} + {x^2}} $	$x = a\tan \theta $	$a\sqrt {1 + {{\tan }^2}\theta } = a\sec \theta $
$\sqrt {{x^2} - {a^2}} $	$x = a\sec \theta $	$a\sqrt {{{\sec }^2}\theta - 1} = a\tan \theta $

Integrating using trigonometric substitution initially produces a function in the angle variable $\theta $. To express the result of the integration into the original variable (variable $x$) the inverse relationship of the trinometric function or arcus function is used.

Examples

Find the following indefinite integrals

1. $\int {\sqrt {64 - {x^2}} } dx$
2. $\int {\frac{{dx}}{{\sqrt {4 - {x^2}} }}} $

Answers

1. For example $x = 8\sin \theta $ then $dx = 8\cos \theta d\theta $
$\begin{array}{l} \int {\sqrt {64 - {x^2}} } dx = \int {\sqrt {{8^2} - {x^2}} } dx\\ = \int {\sqrt {{8^2} - {8^2}{{\sin }^2}\theta } } .8\cos \theta d\theta \\ = \int {8\sqrt {1 - {{\sin }^2}\theta } } .8\cos \theta d\theta \\ = \int {8\sqrt {{{\cos }^2}\theta } } .8\cos \theta d\theta \\ = \int {8\cos \theta } .8\cos \theta d\theta \\ = 64\int {{{\cos }^2}\theta d\theta } \\ = 64\int {\left( {\frac{1}{2} + \frac{1}{2}\cos 2\theta } \right)} d\theta \\ = 64\left( {\frac{1}{2}\theta + \frac{1}{4}\sin 2\theta } \right) + C\\ = 32\theta + 32\sin \theta \cos \theta + C \end{array}$

$x = 8\sin \theta $ then $\sin \theta = \frac{x}{8}$, we get $\theta = \arcsin \left( {\frac{x}{8}} \right)$.

$\sin \theta = \frac{x}{8}$ then $\cos \theta = \frac{{\sqrt {{8^2} - {x^2}} }}{8}$

$\begin{array}{l} \int {\sqrt {64 - {x^2}} } dx = 32\theta + 32\sin \theta \cos \theta + C\\ = 32\arcsin \left( {\frac{x}{8}} \right) + 32.\frac{x}{8}.\frac{{\sqrt {{8^2} - {x^2} } }}{8} + C\\ = 32\arcsin \left( {\frac{x}{8}} \right) + \frac{x}{2}\sqrt {{8^2} - {x^2}} + C \end{array}$

So $\int {\sqrt {64 - {x^2}} } dx = 32\arcsin \left( {\frac{x}{8}} \right) + \frac{x}{2 }\sqrt {{8^2} - {x^2}} + C$

2. For example $x = 2\sin \theta $ then $dx = 2\cos \theta d\theta $
$\begin{array}{l} \int {\frac{{dx}}{{\sqrt {4 - {x^2}} }}} = \int {\frac{{dx}}{{\sqrt {{2^2} - {x ^2}} }}} \\ = \int {\frac{{2\cos \theta d\theta }}{{\sqrt {{2^2} - {2^2}{{\sin }^2}\theta } }}} \\ = \int {\frac{{2\cos \theta d\theta }}{{2\sqrt {1 - {{\sin }^2}\theta } }}} \\ = \int {\frac{{2\cos \theta d\theta }}{{2\cos \theta }}} \\ = \int {d\theta } \\ = \theta + C \end{array}$

$x = 2\sin \theta $ then $\sin \theta = \frac{x}{2}$, we get $\theta = \arcsin \left( {\frac{x}{2}} \right)$

$\begin{array}{l} \int {\frac{{dx}}{{\sqrt {4 - {x^2}} }}} = \theta + C\\ = \arcsin \left( {\frac{x}{2}} \right) + C \end{array}$

So $\int {\frac{{dx}}{{\sqrt {4 - {x^2}} }}} = \arcsin \left( {\frac{x}{2}} \right ) + C$