Answer:
Your process is indeed correct!
The full solution is:
[tex]\displaystyle \int\frac{\sin ^3 x}{\sqrt{\cos x}}\, dx= \frac{2}{5} \cos^{{}^{5}\!/\!{}_{2}} x - 2\cos ^{{}^{1}\!/\!{}_{2}}x + C[/tex]
Step-by-step explanation:
We want to evaluate the integral:
[tex]\displaystyle \int \frac{\sin^3(x)}{\sqrt{\cos(x)}}\, dx[/tex]
As you had done, we can rewrite our integral as:
[tex]\displaystyle =\int \frac{\sin(x)(\sin^2(x))}{\sqrt{\cos(x)}}\, dx[/tex]
Using the Pythagorean Identity, this is:
[tex]\displaystyle =\int \frac{\sin(x)(1-\cos^2(x))}{\sqrt{\cos(x)}}\, dx[/tex]
Now, we can make a substitution. Let u = cos(x). Then:
[tex]\displaystyle du = - \sin x \, dx[/tex]
Substitute:
[tex]\displaystyle = \int\frac{1-u^2}{\sqrt{u}} \, \left(- du\right)[/tex]
Simplify:
[tex]\displaystyle = -\int \frac{1}{\sqrt{u}} - \frac{u^2}{\sqrt{u}}\, du[/tex]
Rewrite:
[tex]\displaystyle = -\int u^{{}^{-1}\!/\!{}_{2}} - u^{{}^{3}\!/\!{}_{2}}\, du[/tex]
By the Reverse Power Rule:
[tex]\displaystyle = -\left(2u^{{}^{1}\!/\!{}_{2}} - \frac{2}{5} u^{{}^{5}\!/\!{}_{2}}\right) + C[/tex]
Simplify:
[tex]\displaystyle = \frac{2}{5} u^{{}^{5}\!/\!{}_{2}}-2u^{{}^{1}\!/\!{}_{2}} + C[/tex]
Back-substitute:
[tex]\displaystyle = \frac{2}{5} \cos^{{}^{5}\!/\!{}_{2}} x - 2\cos ^{{}^{1}\!/\!{}_{2}}x + C[/tex]
