$\int \cos^5 x \sin^{1/2} x \ dx $

Question

$$I=\int \cos^5 x \sin^{1/2} x \ dx $$

My attempt:

$$\cos^5 x = \cos^4 x \cos x =(1-\sin^2x)^2 \cos x$$

\begin{align} I & =\int (1-\sin^2x)^2 \sin^{1/2} x \cos x \ dx \\[10pt] & =\int \sin^{1/2} x \cos x \ dx+\int \sin^2 x \cos x \, dx + \int -2\sin x \cos x\, dx \\[10pt] & ={2\over 3} \sin^{3\over 2} x+{1\over 3} \sin^3x+\cos^2 x+c \end{align}

True ?

Answer 1

You made a tiny mistake with the powers of $\,sin(x)$

$I=\int \cos^5 x \sin^{1/2} x \ dx$

$I =\int (1-\sin^2x)^2 \sin^{1/2} x \cos x \ dx $

let $\,u= sinx \implies du = cos(x)dx$

$I= \int (1-u^2)^2.u^\frac12\,du \\I = \int u^\frac12 + u^{4+\frac12} - 2u^{2+\frac12}\,du\\I = \int u^\frac12+ u^\frac92 - 2u^\frac52\,du\\I = \frac23u^\frac32 + \frac{2}{11}u^\frac{11}{2}-\frac47u^\frac72 +C \\I= \frac23(sin(x))^\frac32+\frac{2}{11}(sin(x))^\frac{11}{2}- \frac47(sin(x))^\frac74 +C$

Answer 2

Your answer somehow seems to be wrong

Hint

Substitute $u=\sin x$ in each integral to simplify to $$\int \sqrt u du-2\int u^{5/2} du+\int u^{9/2} du$$

Which is pretty easy to solve