Integral Reduction for $\sin^7x\cos^3x$ I and using the reduction integral formula for $\sin^7x\cos^3x$
My steps:
First do reduction while $n=7,m=3$, next do a reduction when $n=5,m=3$, next do reduction when $n=3,m=3$, next evaluate integral of $\sin x\cos^3x$. 
After this point I substitute each integral back into the previous equation as you can see from my work and I am unable to see where I am going wrong, it looks correct to me. The correct answer on Wolfram says there is no $\cos$ or $\sin$ value with a degree higher than $1$, it just doesn't make sense to me because the reduction formula shows this as not being the case.

 A: Much too much work, if you want $\int\sin^7\!x\cos^3\!x\,dx$.
When either of the exponents is odd, there’s a supereasy way of handling the problem without using a reduction formula: write $\cos^3\!x=(1-\sin^2\!x)\cos x$, to get
$$
\int(\sin^7\!x-\sin^9\!x)\cos x\,dx\,,
$$
and make the substitution $u=\sin x$, $du=\cos x\,dx$.
A: There can be multiple methods of solving integrals which can yield many different answers which can all differ by a constant.  The easiest way to solve this problem is, as said in my comment,
$$\int\sin^7x\cos^3xdx=\int\sin^7x(1-\sin^2x)\cos xdx=\int(\sin^7x-\sin^9x)\cos xdx=$$
$$\frac18\sin^8x-\frac1{10}\sin^{10}x+C$$
Another possible way to start the integral is through double angle/power reduction identities
$$\int\sin^7x\cos^3xdx=\int(\sin^3x\cos^3x)\sin^4x=\frac1{32}\int\sin^32x(1-\cos2x)^2dx=$$
$$\frac1{32}\int\sin2x(1-\cos^22x)(1+\cos^22x)dx-\frac1{16}\int\sin^32x\cos2xdx=$$
$$-\frac1{64}\cos2x+\frac1{320}\cos^52x-\frac1{128}\sin^42x+C$$
If I have done everything correctly, both answers should be equivalent.  Wolfram verifies both answers differ by $\frac1{80}$.
As for your answer, I was not familiar with the identity that you used, so I looked it up and checked your work.  You might have made things harder by not simplifying fractions as you went, but I can find no mistakes.  So let's try differentiating it to verify.
$$\frac d{dx}\left[\cos^4x(\frac1{10}\sin^6x+\frac3{40}\sin^4x+\frac1{20}\sin^2x-\frac1{40})\right]=$$
$$-4\cos^3x\sin x(\frac1{10}\sin^6x+\frac3{40}\sin^4x+\frac1{20}\sin^2x-\frac1{40})+$$
$$\cos^5x(\frac35\sin^5x+\frac3{10}\sin^3x+\frac1{10}\sin x)=$$
$$\cos^3x(1-\sin^2x)(\frac35\sin^5x+\frac3{10}\sin^3x+\frac1{10}\sin x)$$
$$-\cos^3x(\frac25\sin^7x+\frac3{10}\sin^5x+\frac15\sin^3x-\frac1{10}\sin x)$$
Without going any further, it looks like the sign of the $\sin^7x\cos^3x$ term may be off.  On closer inspection, you might have dropped a minus sign from your reduction formula, assuming wikipedia is correct.
