Integration of $x\sin^6x \cos^4x$

Question

It has been given as, $\int_0^\pi x\sin^6x\cos^4x dx$

I am able to solve $\sin^6x\cos^4x$. But there is a $x$ in between. So could you please help me out.

Answer 1

Using Intuition behind euler's formula,

$$-64\sin^6t=(2i\sin t)^6=(e^{it}-e^{-it})^6=2\cos6t-2\binom61\cos4t+2\binom62\cos2t-2\binom63$$

and $16\cos^4t=(2\cos t)^4=(e^{it}+e^{-it})^4=?$

Now use Werner Formulas, to find the integrals of the form $$\int x\cos mt\ dt$$

and $$\int x\sin mt\ dt$$

Now integrate by parts, $$\int x\cos mt\ dt=x\int\cos mt\ dt-\int\left(\dfrac{dt}{dt}\int\cos mt\ dt\right)dt=?$$

Answer 2

The $x$ in the integrand can be removed easily because the remaining factor behaves very nicely in $[0,\pi]$.

If $f(x)$ is the integrand and $g(x) = f(x) /x$ then we can see that $g(x) =g(\pi-x) $ and hence $$I=\int_{0}^{\pi}xg(x)\,dx=\int_{0}^{\pi}(\pi - x) g(x) \, dx=\pi\int_{0}^{\pi}g(x)\,dx-I$$ so that $$I=\frac{\pi} {2}\int_{0}^{\pi}g(x) \, dx=\pi\int_{0}^{\pi/2}g(x)\,dx=\pi\int_{0}^{\pi/2}\sin^6x\cos^4x\,dx$$ and now use the reduction formula for $$I_{m, n} =\int_{0}^{\pi/2}\sin^m x\cos^n x\, dx$$ given by $$I_{m, n} =\frac{m-1}{m+n} I_{m-2,n}$$ The integral in question is $$I=\pi I_{6,4}=\pi\cdot\frac{5\cdot 3\cdot 1}{10\cdot 8\cdot 6}\cdot\frac{3\cdot 1}{4\cdot 2}\cdot\frac{\pi}{2}=\frac{3\pi^2}{512}$$

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The initial simplification is based on the following very useful identity $$\int_{0}^{a}f(x)\,dx=\int_{0}^{a}f(a-x)\,dx$$ followed by the conditional identity $$\int_{0}^{2a}f(x)\,dx=2\int_{0}^{a}f(x)\,dx$$ which holds provided $f(2a-x)=f(x)$.

The reduction formula for $I_{m, n} $ can be established easily using integration by parts.

Answer 3

Hint: Use integration by parts by setting $u = x$ and $\frac{\mathrm d v}{\mathrm d x} = \sin^6(x)\cos^4(x)$

Answer 4

Hint: $\sin^6x\cos^4x=\frac{1}{32}(1-\cos2x)^3(\cos2x+1)^2=\frac{1}{32}(1-\cos2x)(1-\cos^22x)^2=\frac{1}{32}(1-\cos2x)\sin^22x=\frac{1}{64}(1-\cos2x)(1-\cos4x)=\frac{1}{64}(1-\cos4x-\cos2x+\cos2x\cos4x)=\frac{1}{64}(1-\cos4x-\cos2x+\frac{1}{2}\cos6x+\frac{1}{2}\cos2x)=\frac{1}{64}(1-\cos4x+\frac{1}{2}\cos6x-\frac{1}{2}\cos2x)$. Now multiply by $x$, break it down to multiple integrals and use integration by part for $x\cos kx$ integrals.