Question

Evaluate:$\int {\sin }^4{\mathrm{xcos}}^4\mathrm{xdx}$

Answer 1

To evaluate : $\int {\sin }^4{\mathrm{xcos}}^4\mathrm{xdx}$

Let $\mathrm{I}=\int {\sin }^4\mathrm{x}.{\cos }^4\mathrm{x}.\mathrm{dx}$

We know that

${\sin }^2\mathrm{x}=\frac{(1-\cos 2\mathrm{x})}{2}$ and ${\cos }^2\mathrm{x}=\frac{(1+\cos 2\mathrm{x})}{2}$

Now $\int {\sin }^4{\mathrm{xcos}}^4\mathrm{xdx}$$=$$\int {\sin }^4{\mathrm{xcos}}^4\mathrm{xdx}$

$=$$\int$$\left[{\left(\frac{1-\cos 2\mathrm{x}}{2}\right)}^2{\left(\frac{1+\cos 2\mathrm{x}}{2}\right)}^2\right]\mathrm{dx}$

$=\int {\left(\frac{1-\cos 2\mathrm{x}}{4}\right)}^2{\left(\frac{1+\cos 2\mathrm{x}}{4}\right)}^2]\mathrm{dx}$

$=\frac{1}{16}\int {(1-{\cos }^{2}2\mathrm{x})}^2\mathrm{dx}$

$=\frac{1}{16}\int (1+{\cos }^{4}2\mathrm{x}-2{\cos }^{2}2\mathrm{x})\mathrm{dx}$

Solving for $\cos 2\mathrm{x}$ , we get

$$\begin{gathered} \cos 4\mathrm{x}=2{\cos }^{2}2\mathrm{x}-1 \\ 2{\cos }^{2}2\mathrm{x}=1+\cos 4\mathrm{x} \\ {\cos }^{2}2\mathrm{x}=\frac{1+\cos 4\mathrm{x}}{2} \end{gathered}$$

$\int {\sin }^4{\mathrm{xcos}}^4\mathrm{xdx}$ $=\frac{1}{16}\int (1+{\left(\frac{1+{\cos }^{2}4x}{2}\right)}^2-2(\frac{1+\cos 4x}{2})\mathrm{dx}$

$=\frac{1}{16}\int (1+{\left(\frac{1+{\cos }^{2}4\mathrm{x}}{2}\right)}^2-1-\cos 4\mathrm{xdx}$

$=\frac{1}{16}\int \frac{1+{\cos }^{2}4\mathrm{x}+2\cos 4\mathrm{x}-2\cos 4\mathrm{x}}{2}\mathrm{dx}$

$=\frac{1}{32}\int 1+{\cos }^{2}4\mathrm{xdx}$

$=\frac{1}{32}\int 1+{\cos }^{2}4\mathrm{xdx}$

Solving for $\cos 4x$ , we get

$$\begin{gathered} \cos 8x=2{\cos }^{2}4\mathrm{x}-1 \\ 2{\cos }^{2}4\mathrm{x}=1+\cos 8x \\ {\cos }^{2}4\mathrm{x}=\frac{1+\cos 8x}{2} \end{gathered}$$

$\int {\sin }^4{\mathrm{xcos}}^4\mathrm{xdx}$ $=\frac{1}{32}\int 1+\frac{1+\cos 8\mathrm{x}}{2}\mathrm{dx}$

$=\frac{1}{32}\int \frac{2+1+\cos 8\mathrm{x}}{2}\mathrm{dx}$

$=\frac{1}{64}\int 3+\cos 8\mathrm{x}\mathrm{dx}$

$=\frac{1}{64}[\int 3+\int \cos 8\mathrm{x}]\mathrm{dx}$

$=\frac{1}{64}[3\mathrm{x}+\frac{\sin 8\mathrm{x}}{8}]+\mathrm{C}$ (Formula used : $=\mathrm{cosnx}=\frac{\mathrm{sinnx}}{\mathrm{n}}$)

Thus,$\int {\sin }^4{\mathrm{xcos}}^4\mathrm{xdx}$$=\frac{1}{64}[3\mathrm{x}+\frac{\sin 8\mathrm{x}}{8}]+\mathrm{C}$