# Integration and Proof by induction

My question is as follows: Use induction to prove the following formula for $n \geq 2$

Solution (what I have tried so far):

Base case: $n=2$

RHS: $\int \sin^2xdx = \int 1/2 (1-\cos(2x))dx$

LHS: $\dfrac{-1}{2} \cos x \sin x + \dfrac{2-1}{2} \int dx$

RHS = LHS so base case holds (supposed to be but I haven't worked it out).

Induction Hypothesis (assume true for $n = k$): $\int \sin^kx dx = \dfrac{-1}{k} \cos x(\sin^{k-1}x) + \dfrac{k-1}{k} \int(\sin^{k-2}x)dx$

Induction Step: ($n = k + 1$) requires $\int (\sin^{k+1}x) dx = \dfrac{-1}{k+1} \cos x(\sin^kx) + \dfrac{k}{k+1} \int (\sin^{k-1}x)dx$

So, $\int (\sin^{k+1}x) dx = \in (\sin^kx) dx + \int \sin x dx$

• For the base case using your method of writing $\sin^2(x)$ as $\frac12(1-\cos(2x)$, continue by writing $\int\frac12(1-\cos(2x))dx$ as $\int\frac12dx-\int\cos(2x)dx$. Then the "$\int\frac12dx$" pairs off with the "$\frac{2-1}{2}\int dx$". So you just need to prove that $\int\cos(2x)dx=\frac{-1}2\cos(x)\sin(x)$. To do this, evaluate $\int\cos(2x)dx$ by taking a substitution $u=2x$, then rewrite the resulting answer (which looks like $\sin(2x)$) in terms of $\sin(x)$ and $\cos(x)$. – user1729 Apr 19 '18 at 14:42
• (Of course, you don't need to base case if you use the method in my answer...but anyway...) – user1729 Apr 19 '18 at 14:42

Proof by induction is not necessary here. Suppose $n\geq 2$. First, think of $\sin^n(x)$ as a product $\sin^{n-1}(x)\cdot\sin(x)$. We can then use integration by parts: Take $f'=\sin(x)$ and $g=\sin^{n-1}(x)$ where the rule is $\int f'g=fg-\int fg'$.
This gives the following: \begin{align*} \int\sin^n(x)dx &=-\cos(x)\sin^{n-1}(x)+(n-1)\int\cos^2(x)\sin^{n-2}(x)dx\\ &=-\cos(x)\sin^{n-1}(x)+(n-1)\int(1-\sin^2(x))\sin^{n-2}(x)dx\\ &=-\cos(x)\sin^{n-1}(x)+(n-1)\int\sin^{n-2}(x)dx-k\int\sin^{n}(x)dx\\ \Rightarrow n\int\sin^n(x)dx&=-\cos(x)\sin^{n-1}(x)+(n-1)\int\sin^{n-2}(x)dx\\ \Rightarrow\int\sin^n(x)dx&=-\frac{1}{n}\cos(x)\sin^{n-1}(x)+\frac{n-1}{n}\int\sin^{k-1}(x)dx \end{align*} as required.