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

  • $\begingroup$ Expanding the last term in the fifth equation gives a term $+ \cos 2 x$ that cancels with the $- \cos 2x$ term. $\endgroup$ – Travis Willse Jul 18 '15 at 20:55
  • $\begingroup$ Are you sure ? just x-x=0 , i think $\endgroup$ – Cardinal Jul 18 '15 at 20:56
  • 2
    $\begingroup$ ${ { 1+\cos 4x\over 2}}={1\over 2}+{\cos 4x\over 2}$ and $(1-\sin^2 2x)\cos 2x =\cos 2x-\sin^2 2x\cos 2x$. $\endgroup$ – David Mitra Jul 18 '15 at 20:57
  • $\begingroup$ ah i see ... didnt think of that $\endgroup$ – Panthy Jul 18 '15 at 21:00

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 $$\frac{1}{2} - \frac{1}{2} \cos 4x - \sin^2 2x \cos 2x.$$


Rewriting in a smart way: $$\int\sin^4x\cos^2xdx = \int(\sin^2x)^2\cos^2xdx$$

Half-angle formulas: $$=\int\left(\frac{1-\cos2x}{2}\right)^2\left(\frac{1+\cos2x}{2}\right)dx$$

We use that: $$\begin{align}(1-\cos(2x))^2(1+\cos(2x)) &= (1-\cos(2x))(1-\cos^2(2x)) \\ &= 1-\cos^2(2x) - \cos(2x)+\cos^3(2x)\end{align}$$ to get:$$=\frac18\int[1-\cos2x-\cos^22x+\cos^32x]dx$$

Half-angle formula again along with $\cos^3(2x) = (1-\sin^2(2x))\cos(2x)$ to obtain: $$=\frac18\int\left[\color{red}{1}-\cos2x-\left(\frac{\color{red}{1}+\cos4x}{\color{red}{2}}\right)+(1-\sin^22x)\cos2x\right]dx$$

We use that $1 - \frac{1}{2} = \frac{1}{2}$ to get:

$$=\frac18\int\left[\frac12\color{red}{-\cos 2x}-\frac12\cos 4x+(\color{red}{1}-\sin^22x)\cos2x\right]dx$$

We cancel the $-\cos(2x)$ with the $\cos(2x)$ term to get:



You could also use

$\displaystyle\int\sin^4x\cos^2 x dx=\int(\sin x\cos x)^2\sin^2x dx=\int\big(\frac{1}{2}\sin 2x\big)^2\sin^2x dx$




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