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!
5 Answers
$\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.
$\endgroup$ $\begingroup$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 $n=4,$ $$\int\cos^4xdx=\frac{\cos^3x\sin x}4+\frac34 \int\cos^2xdx$$
$\endgroup$ $\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
$$\text{I}:=\frac{x\cos x\sin x}2-\frac13\cos^3x\sin x+\frac13\text{I}\implies\;\;\ldots$$
$\endgroup$ $\begingroup$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.
$\endgroup$ $\begingroup$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?
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