Evaluate ∫ from 0 to π of sin^3 x dx Step by Step

Step by step

The question

Question: find the integral from 0 to π of sin³ x.

∫ from 0 to π of sin³ x dx =

1 Save one

The power of sin is 3: odd.

Keep one sin x aside.

It will be part of du.

= ∫ from 0 to π of sin² x sin x dx

2 Change the rest

The rule: sin squared plus cos squared is 1.

sin² x + cos² x = 1

So sin² x = 1 − cos² x.

Now everything else is cos x.

= ∫ from 0 to π of (1 − cos² x) sin x dx

3 Choose u

Let u = cos x.

Its derivative is minus sin x: du = −sin x dx.

So sin x dx is −du.

u = cos x, du = −sin x dx

4 Write it with u

cos x becomes u, and sin x dx becomes −du.

= ∫ −(1 − u²) du

Multiply out the bracket.

= ∫ (u² − 1) du

Careful!

The power rule does not work here: the derivative of sin x would have to be there too.

Save one and swap the rest.

Careful: not ∫ sin³ x dx = sin⁴ x4 ✗

5 Integrate

The power rule: add 1 to each power, divide by the new power.

= u³3 − u

6 Put x back

Put cos x back.

No + C: top minus bottom.

= [ cos³ x3 − cos x ] from 0 to π

7 Top minus bottom

The values of cos at the limits.

cos π = −1, cos 0 = 1

Put them in: top minus bottom.

= (23) − (−23)

So the integral is 4/3.

= 43

8 Check

Let's check with a calculator: it adds up 1000 thin strips.

That gives about 1.333, the same as our answer.

Right!

1000 thin strips: ≈ 1.333

Let's look back.

First: save one.

Then: change the rest.

Then: choose u.

Then: write it with u.

Then: integrate.

Then: put x back.

Then: top minus bottom.

Last: check.

Answer