Integration by substitution of sqrt cos theta.sin cube theta

[davie]

Homework Statement

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

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 into the right direction or should I use any other method like integration by parts.

 

[Ivan92]

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

 

[SammyS]

Homework Statement

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

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...

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

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

... and use Ivan92's suggested substitution.

 

[Harrisonized]

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

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

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

 

[davie]

Thanks to everyone who responded, especially Harrisonized, you were right.
Guess I was heading into 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.