Find the indefinite integral of sin^(2n)(x) for a positive integer n. Give a reduction formula, an explicit closed form, and the cases n = 1, 2, 3.

Step-by-step solution

1. 1

   Why a reduction formula rather than a substitution. For an odd power of sine you peel off one factor and substitute $u=\cos x$. That trick fails for an even power $2n$, because no single sine is left over. The standard route is instead integration by parts, which lowers the exponent by two and produces a recursion.

2. 2

   State the reduction formula. Writing $\sin^{m}x=\sin^{m-1}x\cdot\sin x$ and integrating by parts gives the classical result

   $\int\sin^{m}x\,dx=-\frac{\sin^{m-1}x\cos x}{m}+\frac{m-1}{m}\int\sin^{m-2}x\,dx.$

   With $m=2n$ and $I_{n}=\int\sin^{2n}x\,dx$ this becomes

   $I_{n}=-\frac{\sin^{2n-1}x\,\cos x}{2n}+\frac{2n-1}{2n}\,I_{n-1},\qquad I_{0}=\int 1\,dx=x.$

   Each application drops the power by $2$, so after $n$ steps you land on $I_{0}=x$ — which is why an even power always leaves a bare $x$ term in the answer.

3. 3

   Get an explicit closed form from the power-reduction identity. Expanding $\sin^{2n}x$ in cosines of multiple angles,

   $\sin^{2n}x=\frac{1}{2^{2n}}\binom{2n}{n}+\frac{1}{2^{2n-1}}\sum_{k=0}^{n-1}(-1)^{n-k}\binom{2n}{k}\cos\bigl(2(n-k)x\bigr),$

   every term of which integrates instantly. The constant term $\binom{2n}{n}/2^{2n}$ is the mean value of $\sin^{2n}$ over a period, and it is what becomes the $x$ term.

4. 4

   Integrate term by term. Using $\int\cos(\lambda x)\,dx=\frac{\sin(\lambda x)}{\lambda}$ with $\lambda=2(n-k)$:

   $\int\sin^{2n}x\,dx=\frac{\binom{2n}{n}}{2^{2n}}\,x+\frac{1}{2^{2n}}\sum_{k=0}^{n-1}(-1)^{n-k}\binom{2n}{k}\frac{\sin\bigl(2(n-k)x\bigr)}{n-k}+C.$

   The extra factor $\tfrac12$ from $\tfrac{1}{2(n-k)}$ is what turns $2^{2n-1}$ into $2^{2n}$ in the denominator.

5. 5

   Write out the first three cases. Running either formula for $n=1,2,3$:

   $\int\sin^{2}x\,dx=\frac{x}{2}-\frac{\sin 2x}{4}+C,$

   $\int\sin^{4}x\,dx=\frac{3x}{8}-\frac{\sin 2x}{4}+\frac{\sin 4x}{32}+C,$

   $\int\sin^{6}x\,dx=\frac{5x}{16}-\frac{15\sin 2x}{64}+\frac{3\sin 4x}{64}-\frac{\sin 6x}{192}+C.$

   The leading coefficients $\tfrac12,\tfrac38,\tfrac{5}{16}$ are exactly $\binom{2n}{n}/2^{2n}$, matching the Wallis pattern.

6. 6

   Check numerically. Evaluating the closed form on $[0,1.3]$ against Simpson quadrature: $n=1$ gives $0.5211246570$ both ways, $n=2$ gives $0.3310166991$, $n=3$ gives $0.2388166824$. All three agree to ten digits ✓.