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.
Best Answer
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 $u=\sin x$, $du=\cos x\,dx$.