Solutions - 0040-03

(1) Notice all even powers. Use power-to-frequency conversion:

$$\begin{array}{c}{\cos }^{2}x\gg \gg \frac{1}{2}(1+\cos 2x)\\ \\ {\sin }^{2}x\gg \gg \frac{1}{2}(1-\cos 2x)\end{array}$$

Plug in:

$$\int {\sin }^{4}x\cdot {\cos }^{2}xdx\gg \gg \frac{1}{8}\int {(1-\cos 2x)}^2(1+\cos 2x)dx$$

Simplify:

$$\gg \gg \frac{1}{8}\int 1-\cos 2x-{\cos }^2(2x)+{\cos }^3(2x)dx$$

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(2) Reduce power again for ${\cos }^2(2x)$:

$${\cos }^2(2x)\gg \gg \frac{1}{2}(1+\cos (4x))$$

(This is derived from the power-to-frequency formula by changing ‘$x$’ to ‘$2x$’ in that formula.)

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(3) On the last term, swap even bunch:

$${\cos }^3(2x)\gg \gg (1-{\sin }^2(2x))\cos 2x$$

Plug all in and obtain:

$$\gg \gg \frac{1}{8}\int 1-\cos 2x-\frac{1}{2}(1+\cos 4x)+(1-{\sin }^2(2x))\cos 2xdx$$

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(4) Integrate the first three terms:

$$\begin{array}{c}\frac{1}{8}\int 1-\cos 2x-\frac{1}{2}(1+\cos 4x)dx\gg \gg \\ \frac{1}{8}x-\frac{1}{16}\sin 2x-\frac{1}{16}x-\frac{1}{64}\sin 4x+C\end{array}$$

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(5) Integrate the last term with $u$-sub, setting $u=\sin 2x$ and $du=2\cos 2xdx$:

$$\begin{array}{cc}\frac{1}{8}\int (1-{\sin }^2(2x))\cos 2xdx& \gg \gg \frac{1}{16}\int (1-{\sin }^2(2x))2\cos 2xdx\\ & \\ & \gg \gg \frac{1}{16}\int (1-u^2)du\\ & \\ & \gg \gg \frac{u}{16}-\frac{u^3}{48}+C\\ & \\ & \gg \gg \frac{\sin 2x}{16}-\frac{{\sin }^3(2x)}{48}+C\end{array}$$

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(6) Combine in final result:

$$\begin{array}{c}\int {\sin }^{4}x\cdot {\cos }^{2}xdx\gg \gg \\ \frac{1}{16}x-\frac{1}{64}\sin 4x-\frac{1}{48}{\sin }^3(2x)+C\end{array}$$

Note: It is also possible to rewrite ${\sin }^3(2x)$ using trig identities. So, equally valid answers may look different than this.