2
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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
5
$\begingroup$

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

$\endgroup$
1
$\begingroup$

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:

$$=\frac18\int\left[\frac12-\frac12\cos4x-\sin^22x\cos2x\right]\,dx$$

$\endgroup$
0
$\begingroup$

You could also use

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

$\displaystyle=\int\frac{1}{4}\cdot\frac{1}{2}(1-\cos4x)\cdot\frac{1}{2}(1-\cos2x)dx=\frac{1}{16}\int\big(1-\cos4x-\cos2x+\frac{1}{2}(\cos2x+\cos6x)\big)dx$

$\displaystyle=\frac{1}{16}\int\big(1-\cos4x-\frac{1}{2}\cos2x+\frac{1}{2}\cos6x\big)dx=\frac{1}{16}\left[x-\frac{1}{4}\sin4x-\frac{1}{4}\sin2x+\frac{1}{12}\sin6x\right]+C$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.