Integral by Substitution: Definite Integral

How to solve the given definite integral by using the integral by substitution: 2 examples and their solutions.

Example 1

Example

You might want to set sin x = t,
which you've learned before.

Integral by Substitution: Indefinite Integral

The difference is the upper/lower limits.

Let's see how to solve this definite integral
by using the integral by substitution.

Solution

Set sin x = t.

Differentiate both sides.

The derivative of sin x is cos x dx.

And the derivative of t is dt.

cos x dx is already in the given integral.
So you don't have to change cos x dx = dt.

The variable is changed from x to t.

So the upper/lower limits are also changed.

To find the changed limits,
put π/2 and 0
into t = sin x.

If x = 0,
then t = 0.

If x = π/2,
then t = 1.

sin x = t
cos x dx = dt
x = 0 → t = 0
x = π/2 → t = 1

Put these into the given integral:
0π/2 esin x cos x dx.

Then (given) = ∫01 et dt.

Solve the integral.

Definite Integral: How to Solve

The integral of et is itself: et.

Put 1 and 0
into et.

Then e1 - e0.

e1 - e0 = e - 1

So e - 1 is the answer.

Example 2

Example

Solution

Set ln x = t.

Differentiate both sides.

The derivative of ln x is [1/x] dx.

And the derivative of t is dt.

[1/x] dx is already in the given integral.
So you don't have to change [1/x] dx = dt.

The variable is changed from x to t.

So the upper/lower limits are also changed.

To find the changed limits,
put e and e4
into t = ln x.

If x = e,
then t = 1.

If x = e4,
then t = 4.

ln e4 = 4 ln e

Logarithm of a Power

ln x = t
[1/x] dx = dt
x = e → t = 1
x = e4 → t = 4

Put these into the given integral:
ee4 dx/[x ln x].

Then (given) = ∫14 1/t dt.

Solve the integral.

The integral of 1/t is
ln |t|.

Put 4 and 1
into ln |t|.

Then ln |4| - ln |1|.

ln |4| = ln 4 = ln 22

ln |1| = ln 1 = 0

ln 22 = 2 ln 2

So 2 ln 2 is the answer.