5
$\begingroup$

I want to Evaluate integral :$$\int \cos^4(x)\operatorname d\!x$$ And think about doing the following thing: $$ \int \left(1-\sin^2(x)\right)^2\operatorname d\!x \to \int \left(1-2\sin^2(x)+\sin^4(x)\right)\operatorname d\!x $$ but I think I just complicated it.

Any suggestions?
Thanks!

$\endgroup$
  • 2
    $\begingroup$ Hint: en.wikipedia.org/wiki/… $\endgroup$ – vadim123 May 8 '13 at 20:41
  • 2
    $\begingroup$ Bigger hint: $\cos^4x = (\cos^2 x)^2=(\frac{1+\cos 2x}{2})^2$ $\endgroup$ – vadim123 May 8 '13 at 20:42
6
$\begingroup$

$$\cos^4(x) = \left(\dfrac{1+\cos(2x)}2 \right)^2 = \dfrac{1 + \cos^2(2x) + 2\cos(2x)}4 = \dfrac{1 + \dfrac{1+\cos(4x)}2 + 2\cos(2x)}4$$ which gives us $$\cos^4(x) = \dfrac{3 + 4 \cos(2x) + \cos(4x)}8$$ Now you should be able to integrate this off.

$\endgroup$
3
$\begingroup$

Using the reduction formulae,

$$\int\cos^nxdx=\frac{\cos^{n-1}x\sin x}n+\frac{n-1}n \int\cos^{n-2}xdx$$

Putting $n=2,$ $$\int\cos^2xdx=\frac{\cos x\sin x}2+\frac12 \int dx=\frac{\cos x\sin x}2+\frac12 x+C$$

Putting $n=4,$ $$\int\cos^4xdx=\frac{\cos^3x\sin x}4+\frac34 \int\cos^2xdx$$

$\endgroup$
0
$\begingroup$

$$\cos^4x=\cos^2x-\cos^2x\sin^2x\implies$$

$$\implies\text{I}:=\int\cos^4x\,dx=\int\cos^2xdx-\int\cos^2x\sin^2xdx=$$

$$\frac{x+\cos x\sin x}2+\int\sin x\cos^2x\,(-\cos)' xdx$$

Now by parts in the last integral::

$$u=\sin x\;,\;\;u'=\cos x\\v'=\cos^2x\sin x\;,\;\;v=-\frac13\cos^3x$$

so

$$\text{I}:=\frac{x\cos x\sin x}2-\frac13\cos^3x\sin x+\frac13\text{I}\implies\;\;\ldots$$

$\endgroup$
0
$\begingroup$

As indicated in the source pointed out by Vadim123 in the comments,

$\displaystyle \cos^4x=\frac{3 + 4 \cos2x + \cos4x}{8}$

Plug this in and integrate.

$\endgroup$
0
$\begingroup$

One more (using $\cos^4 x = \cos^2 x \cos^2 x = (1-\sin^2 x) \cos^2x$): $$ I=\int \cos^4 x dx = \int \cos^2x dx - \int (\sin x \cos x)^2dx=\int \cos^2 x dx\\ -\frac{1}{4} \int (\sin2x)^2dx=\int \cos^2 x dx-\frac{1}{8}\int \cos^2tdt $$ then use the power reduction formulas: $$ \sin^2 x =\frac{1-\cos2x}{2}\\ \cos^2 x=\frac{1+\cos 2x}{2} $$ Can you handle from here?

$\endgroup$

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.