# Integration by substitution of sqrt cos theta.sin cube theta

1. Dec 30, 2011

### davie

1. The problem statement, all variables and given/known data

To show that $$\int_{0}^ \frac{\pi}{2}\sqrt{cos\theta}sin^3(\theta) d\theta$$ = 8/21

3. The attempt at a solution
The above expression was simplified as
$$\int_{0}^ \frac{\pi}{2}\sqrt{cos\theta}sin^2(\theta) sin(\theta) d\theta$$
$$\int_{0}^ \frac{\pi}{2}\sqrt{cos\theta}(1-cos^2(\theta)) sin\theta d\theta$$
I have tried using integration by substitution method.
Let $$cos\theta = t^2$$
or,$$sin\theta d\theta = 2tdt$$

also changing the limits, when $$\theta = 0$$ , t becomes 1
and when $$\theta = \frac{\pi}{2}$$, t becomes 0

therefore the expression will look like this.
$$\int_{1}^ 0 t.(t^4-1)2t.dt$$

Am I going in to the right direction or should I use any other method like integration by parts.

2. Dec 30, 2011

### Ivan92

Let t = cos θ. See if you can figure it out from there.

3. Dec 30, 2011

### SammyS

Staff Emeritus
Distribute $\sqrt{\cos(\theta)}$ through (1 - cos2(θ)) .

and remember that $\displaystyle \sqrt{x}=x^\frac{1}{2}\,.$

... and use Ivan92's suggested substitution.

4. Dec 31, 2011

### Harrisonized

Why don't you just integrate what you have and find out?

5. Dec 31, 2011

### davie

Thanks to everyone who responded, especially Harrisonized, you were right.
Guess I was heading in to the right path.
$$\int_{1}^ 0 t(t^4-1)2t dt$$

--->$$2.\frac{t^7}{7}-2.\frac{t^3}{3}$$ when integrated.

--->$$0-\frac{6-14}{21}$$ when variable substituted with the limits.