Integration Techniques: Solving Definite Integrals using Substitution, Study notes of Mathematics

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Integration Techniques

Example

Integrate

∫ x^3 ln(x)dx

1

A solution

Let u = x^4 so that du = 4x^3 dx. Note that 4 ln(x) = ln(x^4 ). So,

∫ x^3 ln(x)dx = 1 16

ln(x^4 )(4x^3 )dx

= 1 16

ln(u)du

= 16 1 (u ln(u) − u) + C

=

4 x

(^4) ln(x) − 1 16 x

4 + C

Another solution

Set u = ln(x) and dv = x^3 dx. So that du = dxx and v = 14 x^4.

Then, ∫ x^3 ln(x)dx =

udv

= uv −

vdu

= 1 4

x^4 ln(x) −

x^4 x

dx

= 1 4

x^4 ln(x) − 1 4

x^3 dx

= 14 x^4 ln(x) − 161 x^4 + C

3

Another example

Compute

∫ (^4) x √ x − 1

dx

Solution

Set u = ln(ln(x)) so that du = (^) x ln(dxx).

Then

∫ (^) ln(ln(x)) x ln(x) dx^ =

du = u + C = ln(ln(x)) + C

7

Another example

Compute

∫ (^) π

−π

x sin(x^4 )dx

Solution

If ∫ f (x) = x sin(x^4 ), then f (−x) = −f (x). Thus, 0 −π x^ sin(x

(^4) )dx = − ∫^ π 0 x^ sin(x

(^4) )dx. So, ∫^ π −π x^ sin(x

(^4) )dx = 0.