Q.Evaluate : $\displaystyle\int \sin^6 x\, dx$.

The substitution $x=a\sin\theta$ turns the integral into Wallis' formula for $\int_0^{\pi/2}\cos^4\theta\,d\theta$.

1. Let $x=a\sin\theta$, so $dx=a\cos\theta\,d\theta$; as $x:0\to a$, $\theta:0\to\pi/2$, and $a^2-x^2=a^2\cos^2\theta$.
2. $\left(\sqrt{a^2-x^2}\right)^3=(a^2-x^2)^{3/2}=a^3\cos^3\theta$.
3. $\displaystyle\int_0^a(a^2-x^2)^{3/2}dx=\int_0^{\pi/2}a^3\cos^3\theta\cdot a\cos\theta\,d\theta=a^4\int_0^{\pi/2}\cos^4\theta\,d\theta$. …

Using the symmetry of $\sin^4x$ about $x=\pi/2$ and Wallis' reduction formula, $\displaystyle\int_0^{\pi}\sin^4x\,dx=\dfrac{3\pi}{8}$.

1. Since $\sin(\pi-x)=\sin x$, the graph of $\sin^4x$ on $[0,\pi]$ is symmetric about $x=\pi/2$, so $\displaystyle\int_0^{\pi}\sin^4x\,dx = 2\int_0^{\pi/2}\sin^4x\,dx$.
2. By Wallis' (reduction) formula, for even $n$, $\displaystyle\int_0^{\pi/2}\sin^n x\,dx=\dfrac{(n-1)(n-3)\cdots 1}{n(n-2)\cdots 2}\cdot\dfrac{\pi}{2}$. …

Using the symmetry $\sin^4(\pi-x)=\sin^4x$ together with Wallis' reduction formula for $\int_0^{\pi/2}\sin^nx\,dx$ gives $\int_0^\pi\sin^4x\,dx=\dfrac{3\pi}{8}$.

1. Note that $\sin(\pi - x) = \sin x$, so $\sin^4(\pi-x) = \sin^4 x$; the graph of $\sin^4x$ on $[0,\pi]$ is symmetric about $x=\pi/2$.
2. Hence $\displaystyle\int_0^\pi \sin^4x\,dx = 2\int_0^{\pi/2}\sin^4x\,dx$. …

Substituting $u=2x$ and using the standard Wallis result $\int_0^{\pi/2}\cos^3u\,du=\dfrac23$ gives the integral value $\dfrac13$.

1. Let $u=2x\Rightarrow du=2\,dx$; when $x=0,\ u=0$, and when $x=\pi/4,\ u=\pi/2$.
2. $\displaystyle\int_0^{\pi/4}\cos^32x\,dx=\dfrac12\int_0^{\pi/2}\cos^3u\,du$. …

Derive the reduction formula for $\int\cos^n x\,dx$ using integration by parts, splitting $\cos^n x=\cos^{n-1}x\cdot\cos x$; the result is the standard textbook formula with a $+$ sign and coefficient $\frac1n$.

1. Write $I_n = \displaystyle\int \cos^{n-1}x\cdot\cos x\,dx$.
2. Integrate by parts with $u=\cos^{n-1}x$ (so $du=-(n-1)\cos^{n-2}x\sin x\,dx$) and $dv=\cos x\,dx$ (so $v=\sin x$): $I_n = \cos^{n-1}x\sin x - \int \sin x\cdot[-(n-1)\cos^{n-2}x\sin x]\,dx$
3. $I_n = \cos^{n-1}x\sin x + (n-1)\int \cos^{n-2}x\sin^2x\,dx$
4. Use $\sin^2x = 1-\cos^2x$: $I_n = \cos^{n-1}x\sin x + (n-1)\int\cos^{n-2}x\,dx - (n-1)\int\cos^n x\,dx$
5. So $I_n = \cos^{n-1}x\sin x + (n-1)I_{n-2} - (n-1)I_n$. …