Compute the integral using trig substitution

[imattxc]

I need to correct the following problem using trig substitution.



I keep getting stuck at:



If I dont use trig sub and just use u-sub I get like 2/15, but I think I need to answer using trig sub. Thanks for any help you can offer!

 

[Deleted member 4993]

I need to correct the following problem using trig substitution.
View attachment 4649


I keep getting stuck at:

View attachment 4648

If I dont use trig sub and just use u-sub I get like 2/15, but I think I need to answer using trig sub. Thanks for any help you can offer!

sin³(Θ) * cos²(Θ) = sin(Θ) * [1-cos²(Θ)] * cos²(Θ) = sin(Θ) * cos²(Θ) - sin(Θ) * cos⁴(Θ) .... Now integrate....

 

[HallsofIvy]

Typically, if you have an odd power of sine or cosine you can factor one out, to use with the "dx", then use $\displaystyle sin^2(x)+ cos^2(x) = 1$, so that either $\displaystyle sin^2(x)= 1- cos^2(x)$ or $\displaystyle cos^2(x)= 1- sin^2(x)$ to reduce the remaining even power. Here, you have $\displaystyle \int_0^{\pi/2} sin^3(\theta) cos^2(\theta)d\theta$ which we can write as $\displaystyle \int_0^{\pi/2} sin^2(\theta) cos^2(\theta) (sin(\theta) d\theta)= \int_0^{\pi/2} (1- cos^2(\theta))cos^2(\theta)(sin(\theta)d\theta)= \int_0^{\pi/2} (cos^2(\theta)- cos^4(\theta))(sin(\theta)d\theta)$.

Now make the substitution $\displaystyle u= cos(\theta)$.