Use integration by parts to prove the reduction formula $$\int\sin^n(x)\ dx = – {\sin^{n-1}(x)\cos(x)\over n}+{n-1\over n}\int\sin^{n-2}(x)\ dx$$
So I'm definitely on the right track because I'm very close to this result, and I also found an example of this exact question in one of my textbooks.
I made f(x)=$\sin^{n-1}(x)$ and g'(x)=$\sin(x)$. And with that I got to the point of having
$\int\sin^n(x)\ dx = – \sin^{n-1}(x)\cos(x)+(n-1)\int\cos^2(x)\sin^{n-2}(x)\ dx$
(sorry I haven't shown how I got to this stage, It's taking me way too long to write out these formulas, but the textbook done the same thing)
Then the textbook then uses the identity $\cos^2(x)=1-\sin^2(x)$.
Which would give:
$\int\sin^n(x)\ dx = – \sin^{n-1}(x)\cos(x)+(n-1)\int(1-\sin^2(x))\sin^{n-2}(x)\ dx$
And then I would have thought expanding that last integral would give
$\int\sin^n(x)\ dx = – \sin^{n-1}(x)\cos(x)+(n-1)\int\sin^{n-2}(x)-\int\sin^n(x))\\ dx$
However the textbook says you should get
$\int\sin^n(x)\ dx = – \sin^{n-1}(x)\cos(x)+(n-1)\int\sin^{n-2}(x)-(n-1)\int\sin^n(x))\\ dx$
Lots of information but essentially my question is just this, where did that extra (n-1) come from? I can't figure it out and its preventing me from finishing the question.
Thanks in advance 🙂
Best Answer
the $n-1$ term is multiplying everything against the integral on the right hand side; namely the term \begin{equation} (n-1)\left(\int(1-\sin^2(x))\sin^{n-2}(x)dx\right) \end{equation} As such once multiplied out; the $n-1$ term appears agains the $-\int \sin^{n}(x)dx$ term too.