I am reading through my textbook and there is a part of the solution to an example that I do not understand…
$$\int\sin^4x\cos^2x\,dx = \int(\sin^2x)^2\cos^2x\,dx$$
$$=\int\left(\frac{1-\cos2x}{2}\right)^2\left(\frac{1+\cos2x}{2}\right)\,dx$$
$$=\frac18\int[1-\cos2x-\cos^22x+\cos^32x]\,dx$$
$$=\frac18\int\left[1-\cos2x-\left(\frac{1+\cos4x}{2}\right)+(1-\sin^22x)\cos2x\right]\,dx$$
$$=\frac18\int\left[\frac12-\cos2x-\frac12\cos4x+(1-\sin^22x)\cos2x\right]\,dx$$
$$=\frac18\int\left[\frac12-\frac12\cos4x-\sin^22x\cos2x\right]\,dx$$
I really don't understand what they did after the 5th equation, for example how the $1$ became $\frac12$. If anyone can explain the algebra to me from the 5th part that would be great… thanks..
EDIT: I don't care about the final answer.. just wondering how they transitioned to the next few steps
Best Answer
From this, we break the fraction into its individual components so that we go from $$1-\cos2x-\left(\frac{1+\cos4x}{2}\right)+(1-\sin^22x)\cos2x$$
to $$\color{red}{1} - \cos 2x \color{red}{- \frac{1}{2}} - \frac{1}{2}\cos 4x + (1-\sin^2 x)\cos 2x$$
Simplifying gives us
$$\frac12-\cos2x-\frac12\cos4x+(1-\sin^22x)\cos2x$$ They expanded the last bracket to get $$\frac12 \color{blue}{ - \cos2x}-\frac12 \cos4x+ \color{blue}{\cos 2x}-\sin^22x\cos2x$$ We can re-arrange this a little to get $$\frac{1}{2} - \cos 2x + \cos 2x - \frac{1}{2} \cos 4x - \sin^2 2x \cos 2x$$
Notice how $\cos 2x - \cos 2x = 0$ to get $$\frac{1}{2} - \frac{1}{2} \cos 4x - \sin^2 2x \cos 2x.$$