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Integration by substitution

Integration by substitution

In Calculus, Integration by substitution is a method where we replace an integral value. We replace its parts to make the expression easier to integrate.

Consider the function f(u) such that u is a function of x.

let u = f(x) and f(u)=u

$$\frac{d}{dx} f(u) = \frac{du}{dx} f'(u)$$

where:

$$f'(u) = \frac{df(u)}{du}$$

Integrating….;

$$\int \frac{du}{dx}f'(x) =\int \frac{d}{dx}f(u)dx = f(u) + C$$

remember that:

$$ f'(u) =\frac{df(u)}{du}$$
$$\int f'(u)du = f(u) + c$$
$$\int \frac{du}{dx}f'(u)dx = \int f'(u)du = f(u) + c$$
integration by substition

Example 1

Solve the problem below using Integration by substitution:

$$\int x^2 \sqrt{x^3 +5}$$

solution

substitute the expression under the root sign as shown:

let

u=x3+5u

And then integrating u with respect to x we get:

$$\frac{du}{dx} = 3x^2$$

and hence:

du=3x2dx

and making dx the subject:

$$dx = \frac{du}{3x^2}$$

rewriting the expression:

$$x^2 \sqrt{x^3+5} = \int x^2 \sqrt{u} \frac{du}{3x^2}$$

x2 on the numerator can then be canceled out by the x2 on the denominator.

The expression becomes:

$$\int \frac{\sqrt{u}du}{3x^2} = \int \frac{u^{\frac{1}{2}}}{3}du=\frac{1}{3} \int U^{\frac{1}{2}} du$$
$$\frac{1}{3}.\frac{U^{\frac{3}{2}}}{\frac{3}{2}} + c = \frac{2U^{\frac{3}{2}}}{9} + c $$
$$\frac{1}{3}.\frac{U^{\frac{3}{2}}}{\frac{3}{2}} + c = \frac{2U^{\frac{3}{2}}}{9} + c = \frac{2}{9}{(x^3 +5)}^{\frac{3}{2}} $$

Example 2

solve :

$$\int 2x \sqrt{1 + x^2}dx$$

solution

let u = 1 +x2

then differentiating u with respect to x

$$\frac{du}{dx}=2u$$

and then rearranging:

$$dx = \frac{du}{2x}$$

then rewriting the integral:

$$\int 2x \sqrt{u} \frac{du}{2x} = \int \sqrt{u}du$$

please note that, 2x on the denominator has canceled the 2x on the numerator so that we only have u that is easy to integrate:

$$\int \sqrt{u}du = \int u^{\frac{1}{2}}du = \left [\frac{u^{3}{2}}{\frac{3}{2}}+c \right]$$
$$\frac{2u^{\frac{3}{2}}}{3}+c = \frac{2(1+x^2)^{\frac{3}{2}}}{3} + c$$

Example

solve by substitution:

$$\int \frac{1}{\sqrt{2x+1}}dx $$

solution

let u = 2x + 1

differentiating u with respect to x we obtain;

$$\frac{du}{dx}=2$$

now we express dx in terms of u as:

$$du = \frac{dx}{2}$$

Then we need to rewrite our integral as:

$$\int \frac{1}{u^{\frac{1}{2}}}.\frac{du}{2} = \frac{1}{2} \int u^{\frac{-1}{2}}du$$

then:

$$\int u^{\frac{-1}{2}}du = \frac{1}{2} \left[ \frac{u^{\frac{1}{2}}}{\frac{1}{2}} \right] +c = U^{\frac{1}{2}} +c $$

rem U = 2x + 1

and any number raised to power of 1/2 is like finding it’s squareroot

hence:

$$u^{\frac{1}{2}}+c = \sqrt{2x +1}+c$$

substitution Problem involving trigonometric ratios

solve: ∫cosxsin3xdx

solution:

let u = sin x: then

$$\frac{du}{dx} = cos \ x$$
$$dx = \frac{du}{cos \ x}$$

and applying substitution in our integral, we replace dx with du;

$$cos \ x \ sin^3 \ x d x = \int cos \ x . u^3 \ \frac{du}{cos \ x }$$

cos x in the numerator cancels the one at the denominator so that the expression is reduced to:

$$\int u^3 du = \frac{U^4}{4}+c$$

but u = sin x; therefore:

$$\frac{u^4}{4}+c = \frac{sin^4 \ x}{4}+c$$

generally:

$$\int cos \ x sin^n x dx = \frac{1}{n+1}sin^{n+1}+c$$

Example2 on trigonometric substitution

solve:

$$sin \ x cos^5 \ x dx$$

solution

let u = cos x

$$du = -sinx \ dx$$ $$\frac{du}{-sinx}=dx$$

rewriting dx in our original integral:

$$\int sinxcos^5xdx = \int sinxcos^5x \frac{du}{-sinx} $$

we eliminate sin x in the integral to obtain the following expression:

$$\int -cos^5xdu$$

but cos x = u

hence cos5x = u5 and so the integral becomes:

$$- \int u^5du$$

hence we get:

$$- \int u^5du = – \frac{u^6}{6}+c$$

and then replacing u for cos x we get:

$$- \frac{cos^6x}{6}+c$$

in general:

$$\int sinx \ cos x dx = \frac{-1}{n+1}cos^{n+1}x+c$$

Integration by substitution: natural log by substitution

Integration by substitution, specifically with natural logarithms, involves choosing a part of the integrand as a new variable. This is usually denoted as ‘u’ which strategically simplifies the integration process.

The natural logarithm often appears in the denominator of a fraction where the numerator is its derivative, making it suitable for substitution.

General Steps for U-Substitution with Natural Logarithms are as follow:

1. Find a suitable ‘u’.

look for a part of the integrand where its derivative is also present in the integrand. This Often will be the natural logarithm term itself, or a function containing the natural logarithm.

2. Find du:

First, calculate the derivative of your chosen ‘u’ with respect to the original variable, usually ‘x’. Then, express it in the form du = (derivative of u) dx.

3. Substitute and Simplify

Replace the original expression in the integral with ‘u’ and ‘du’. The integral should now be in a simpler form, involving a basic integration rule.

4. Solve the new integral with respect to ‘u’.

5. Substitute

Replace ‘u’ with its original expression in terms of the original variable to get the final result.

6. Add the constant of integration ‘c’:

Since you are solving an indefinite integral, don’t forget to add the constant of integration ‘+ c’. 

Example:

solve:

$$\frac{ln \ x}{x}dx$$

solution

let u = ln x

$$\frac{du}{dx}=\frac{1}{x}$$

x du = dx

hence we write the integral as:

$$\int \frac{ln \ x}{x}dx =\int \frac{ln \ x}{x}xdu$$

and so we eliminate x to have the integral:

$$\int ln x \ du = \int du$$

∫udu=u22+C=(lnx)22+C∫udu=u22+C=(lnx)22+C

$$udu=\frac{u^2}{2} + c = \frac{(lnx)^2}{2}n+c$$

Example on secant trigonometric integration

solve:

$$\int sec^2(3x-1)dx$$

solution

let u = 3x-1

$$\frac{du}{dx}=3$$ $$du= 3dx$$ $$dx = \frac{du}{3}$$
$$\int sec^2u\frac{du}{3} = \frac{1}{3} \int sec^2u du = \frac{1}{3}tanu+c=\frac{1}{3}tan(3x-1)+c$$

Example 3 on trigonometric integration by substitution

integrate:

$$sinx \ e^{cosx}dx$$

solution

$$ let \ u = cos \ x$$ $$du = -sinx dx$$
$$dx = \frac{du}{-sinx}$$
$$=\int sinxe^u \frac{du}{-sinx}=- \int e^udu = -e^u +c$$ $$$$

Practice exercise

use the substitution method to evaluate the following integrals:

$$1. \ \int x(x^2-3)^4 dx$$ $$2. \ \int x\sqrt{1-x^2 }dx$$ $$3. \ \int cos2x(sin(2x+3))^2dx$$ $$4. \ \int e^x \sqrt{1+e^x}dx$$ $$5. \ \int sec^2x tan^2xdx$$ $$6. \ \int \sqrt{x} \sqrt{1+x^{\frac{3}{2}}}dx$$ $$7. \ \int (x+1)sin(x^2 +2x +2)dx$$ $$8. \ \int (x+1) \sqrt{x^2+2x}dx$$ $$9. \ \int \frac{x^2}{(x^3+8)^4}dx$$

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