How do you find the integral of $\int \sin^{5/2}x \cos^3x$

$\int\sin^{5/2}x\cos^3xdx$

I'm confused because I don't know if I should square the entire thing and get

$\int \sin^5x \cos^6xdx$ and then break up $\sin^5x=\sin^4x \sin x$

But does the $dx$ get squared too? How would I go about solving this?

Solution 1:

$$ \begin{aligned} \int{\sin^{\frac{5}{2}}(x)\cos^3(x)dx} &= \int{\sin^{\frac{5}{2}}(x)\cos^2(x)\cos(x)dx} = \int{\sin^{\frac{5}{2}}(x)\left(1-\sin^2(x)\right)\cos(x)dx} = \\ &= \left| \begin{aligned} u &= \sin(x) \\ du &= \cos(x)dx \end{aligned} \right| = \int{u^\frac{5}{2}\left(1-u^2\right)du} = \int{u^\frac{5}{2}du}-\int{u^\frac{9}{2}du} \end{aligned} $$

Now, you have to apply the formula $$ \int{u^a}du = \frac{u^{a+1}}{a+1}+C $$

to both integrals.

Solution 2:

Recalling that $$\frac{d}{dx}[\sin x] = \cos x,$$ it seems natural to want to choose a substitution of the form $$u = \sin x, \quad du = \cos x \, dx.$$ But you have $\cos^3 x$ in the integrand, not $\cos x$. What trigonometric identities can you apply to change $\cos^3 x$ into some function of $\sin x$ multiplied by $\cos x$; i.e., how can we write $\sin^{5/2} x \cos^3 x$ in the form $$f(\sin x) \cdot \cos x$$ for some suitable function $f$?