Integration by substitution

illustrating integration by substitution by replacing u for x-1

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In Calculus, Integration by substitution is when we replace an integral value or its parts in an attempt 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

ddxf(u)=dudxf′(u)ddxf(u)=dudxf′(u)

where:

f′(u)=df(u)duf′(u)=df(u)du

Integrating….;

∫dudxf′(u)dx=∫ddxf(u)dx=f(u)+C∫dudxf′(u)dx=∫ddxf(u)dx=f(u)+C

but:

f′(u)=df(u)duf′(u)=df(u)du

so that

∫f′(u)du=f(u)+c∫f′(u)du=f(u)+c

∫dudxf′(u)dx=∫f′(u)du=f(u)+C∫dudxf′(u)dx=∫f′(u)du=f(u)+C

Example 1

Solve the problem below using Integration by substitution:

∫x2√x3+5∫x2x3+5

solution

substitute the expression under the root sign as shown:

let

u=x3+5u=x3+5

then integrating u with respect to x we get:

dudx=3x2dudx=3×2

and so:

du=3x2dxdu=3x2dx

and making dx the subject:

dx=du3x2dx=du3x2

rewriting the expression:

∫x2√x3+5=∫x2√udu3x2∫x2x3+5=∫x2udu3x2

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

The expression becomes:

∫√Udu3x2=∫u123du∫Udu3x2=∫u123du

=13∫U12du=13∫U12du

13.U3232+C=2U329+C13.U3232+C=2U329+C

13.U3232+C=2U329+C=29(x3+5)32+C13.U3232+C=2U329+C=29(x3+5)32+C

Example 2

solve :

∫2x√1+x2dx∫2×1+x2dx

solution

let u = 1 +x²

then differentiating u with respect to x

dudx=2ududx=2u

and then rearranging:

dx=du2xdx=du2x

then rewriting the integral:

∫2x√udu2x=∫√udu∫2xudu2x=∫udu

2x on the denominator has cancelled the 2x on the numerator so that we only have u that is easy to integrate:

∫√udu=∫u−12du=[u3232+C]∫udu=∫u−12du=[u3232+C]

=2u323+C=2(1+x2)323+C=2u323+C=2(1+x2)323+C

Example

solve by substitution:

∫1√2x+1dx∫12x+1dx

solution

let u = 2x + 1

differentiating u with respect to x we obtain;

dudx=2dudx=2

now we express dx in terms of u as:

du=dx2du=dx2

Then we need to rewrite our integral as:

∫1u12du2=12∫u−12du∫1u12du2=12∫u−12du

then:

∫u−12du=12[u1212]+C∫u−12du=12[u1212]+C

=U12+C=U12+C

rem U = 2x + 1

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

hence:

U12+C=√2x+1+CU12+C=2x+1+C

substitution Problem involving trigonometric ratios

solve:

∫cosxsin3xdx∫cosxsin3xdx

solution:

let u = sin x: then

dudx=cosxdudx=cosx

dx=ducosxdx=ducosx

and applying substitution in our integral, we replace dx;

∫cosxsin3xdx=∫cosx.u3ducosx∫cosxsin3xdx=∫cosx.u3ducosx

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

∫u3du∫u3du

∫u3du=U44+C∫u3du=U44+C

but u = sin x; therefore:

U44+C=sin4x4+CU44+C=sin4x4+C

generally:

∫cosxsinxxdx=1n+1sinn+1+C∫cosxsinxxdx=1n+1sinn+1+C

Example on trigonometric substitution

solve:

∫sinxcos5xdx∫sinxcos5xdx

solution

let u = cos x

du=−sinxdxdu=−sinxdx

du−sinx=dxdu−sinx=dx

rewriting dx in our original integral:

∫sinxcos5xdx=∫sinxcos5xdu−sinx∫sinxcos5xdx=∫sinxcos5xdu−sinx

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

=∫−cos5xdu=∫−cos5xdu

but cos x = u

hence cos⁵x = u⁵ and so the integral becomes:

−∫u5du−∫u5du

hence we get:

−∫u5du=−u66+6−∫u5du=−u66+6

and then replacing u for cos x we get:

−cos6x6+C−cos6x6+C

in general:

∫sinxcosnxdx=−1n+1cosn+1x+C∫sinxcosnxdx=−1n+1cosn+1x+C

Integration by substitution: natural log by substitution

solve:

∫lnxxdx∫lnxxdx

solution

let u = lnx

dudx=1xdudx=1x

x du = dx

hence we write the integral as:

∫lnxxdx=∫lnxxxdu∫lnxxdx=∫lnxxxdu

and so we eliminate x to have the integral:

∫lnxdu=∫udu∫lnxdu=∫udu

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

Example on sec trigonometric ration

solve:

∫sec2(3x−1)dx∫sec2(3x−1)dx

solution

let u = 3x-1

dudx=3dudx=3

du = 3dx

dx=du3dx=du3

∫sec2udu3=13∫sec2udu=13tanu+c=13tan(3x−1)+C∫sec2udu3=13∫sec2udu=13tanu+c=13tan(3x−1)+C

Example 3 on trigonometric ratio

integrate:

∫sinxecosxdx∫sinxecosxdx

solution

let u = cos x

du=−sinxdxdu=−sinxdx

dx=du−sinxdx=du−sinx

=∫sinxeudu−sinx=−∫eudu=−eu+c=∫sinxeudu−sinx=−∫eudu=−eu+c

Practice exercise

use the substitution method to evaluate the following integrals:

1.∫x(x2−3)4dx∫x(x2−3)4dx

2.∫x√1−x2dx∫x1−x2dx

3.∫cos2x(sin(2x+3))2dx∫cos2x(sin(2x+3))2dx

4.∫ex√1+exdx∫ex1+exdx

5.∫sec2xtan2xdx∫sec2xtan2xdx

6.∫√x√1+x32dx∫x1+x32dx

7.∫(x+1)sin(x2+2x+2)dx∫(x+1)sin(x2+2x+2)dx

8.∫(x+1)√x2+2x+3dx∫(x+1)x2+2x+3dx

9.∫x2(x3+8)4dx∫x2(x3+8)4dx

Tag: Integration by substitution

Example 1

solution

Example 2

solution

Example

solution

substitution Problem involving trigonometric ratios

Example on trigonometric substitution

Integration by substitution: natural log by substitution

Example on sec trigonometric ration

Example 3 on trigonometric ratio

Practice exercise

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