Compute the integral using trig substitution

imattxc

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I need to correct the following problem using trig substitution.
eqn4315.png


I keep getting stuck at:

eqn4315.png

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!
 
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....
 
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 \(\displaystyle u= cos(\theta)\).
 
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