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.