Kathleen Rausch

2021-12-27

Let the integral be:
$\int {\mathrm{sin}}^{4}\left(x\right){\mathrm{cos}}^{3}\left(x\right)dx$
I have to integrate this function by changing the variable. I'm trying: $u=\mathrm{sin}\left(x\right)$ and so $du=\mathrm{cos}\left(x\right)dx$. By rewriting the integral I get:
$\int {u}^{4}{\mathrm{cos}}^{2}\left(x\right)du$
But I'm stuck here because I'm not sure there should be any expression with x left in the integral.Also I know the final answer is :
$\int {\mathrm{sin}}^{4}x{\mathrm{cos}}^{3}xdx=-\frac{{\mathrm{sin}}^{7}x}{7}+\frac{{\mathrm{sin}}^{5}x}{5}$

Samantha Brown

Hint:
Use the identity
${\mathrm{cos}}^{2}\left(x\right)=1-{\mathrm{sin}}^{2}\left(x\right)$
and don't forget the constant of integration!

Toni Scott

As the exponent of cos is odd, you can set
$u=\mathrm{sin}x,du=\mathrm{cos}xdx$
If the function to integrate had been, say ${\mathrm{sin}}^{3}x{\mathrm{cos}}^{4}xdx$, we would have set
$u=\mathrm{cos}x,du=-\mathrm{sin}xdx$

Vasquez

$\int {\mathrm{sin}}^{4}x{\mathrm{cos}}^{3}xdx=-\int -\frac{1}{5}{\mathrm{cos}}^{2}x\left(-5\mathrm{cos}\left(x\right){\mathrm{sin}}^{4}x\right)dx$
$=-\int -\frac{1}{5}{\mathrm{cos}}^{2}x\left({\mathrm{sin}}^{5}x{\right)}^{\prime }dx$
Now perform an integration by parts and use the same technique for the new integral.

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