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 :

$\int {\mathrm{sin}}^{4}x{\mathrm{cos}}^{3}xdx=-\frac{{\mathrm{sin}}^{7}x}{7}+\frac{{\mathrm{sin}}^{5}x}{5}$

I have to integrate this function by changing the variable. I'm trying:

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 :

Samantha Brown

Beginner2021-12-28Added 35 answers

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!

Use the identity

and don't forget the constant of integration!

Toni Scott

Beginner2021-12-29Added 32 answers

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

$u=\mathrm{cos}x,du=-\mathrm{sin}xdx$

If the function to integrate had been, say

Vasquez

Expert2022-01-09Added 669 answers

Now perform an integration by parts and use the same technique for the new integral.

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