Nihilist Comedian
Jun11-04, 10:10 PM
\int{\sin^{n}(x)\cos^{m}(x)dx}
=\frac{\sin^{n+1}(x)\cos^{m-1}(x)}{n+1}+\frac{m-1}{n+1}\int{\sin^{n+2}(x)\cos^{m-2}(x)dx}
That was quite easy, but it's the simplification process following this that throws me. My answer is perfectly correct, but it is simplified in the answers (in my maths book) to the following form.
\frac{\sin^{n+1}(x)\cos^{m-1}(x)}{n+m}+\frac{m-1}{n+m}\int{\sin^{n}(x)\cos^{m-2}(x)dx}
The form that I had it in can be used to calculate integrals for specific values of n an m, though in an exam, I believe that I'd have to express it in simpler form to get full marks.
Thanks for the help.
=\frac{\sin^{n+1}(x)\cos^{m-1}(x)}{n+1}+\frac{m-1}{n+1}\int{\sin^{n+2}(x)\cos^{m-2}(x)dx}
That was quite easy, but it's the simplification process following this that throws me. My answer is perfectly correct, but it is simplified in the answers (in my maths book) to the following form.
\frac{\sin^{n+1}(x)\cos^{m-1}(x)}{n+m}+\frac{m-1}{n+m}\int{\sin^{n}(x)\cos^{m-2}(x)dx}
The form that I had it in can be used to calculate integrals for specific values of n an m, though in an exam, I believe that I'd have to express it in simpler form to get full marks.
Thanks for the help.