Question

Find the integrals of the function :
${\cos }^{4}2x$

Answer 1

Given,

$\Rightarrow {\cos }^{4}2x$

Others trigonometry ratios relation are below

${\left({\cos }^{2}2x\right)}^2={\left(\frac{\cos 2x+1}{2}\right)}^2={\left(\frac{1-\cos 2x}{2}\right)}^2$

We know that,

$\because \cos 2A=2{\cos }^{2}A-1=1-2{\sin }^{2}A=\frac{1-{\tan }^{2}A}{1+{\tan }^{2}A}$

Therefore,

${\cos }^{4}2x={\left(\cos 2x\right)}^4={\left(\frac{1-{\tan }^{2}x}{1+{\tan }^{2}x}\right)}^4$

${\cos }^{4}2x={\left(\cos 2x\right)}^4={\left(\frac{1}{\sec x}\right)}^4$

${\cos }^{4}2x={\left(\cos 2x\right)}^4={\left(\frac{1}{\sec x}\right)}^4={\left(\frac{1}{{\sec }^{2}x}\right)}^2$

$={\left(\frac{1}{1+{\tan }^{2}x}\right)}^2={\left(\frac{1}{1+\frac{1}{{\cot }^{2}x}}\right)}^2={\left(\frac{{\cot }^{2}x}{{\cot }^{2}x+1}\right)}^2$

${\cos }^{4}2x={\left(\frac{{\cot }^{2}x}{{\cot }^{2}x+1}\right)}^2={\left(\frac{\cos e{c}^{2}x-1}{\cos e{c}^{2}x}\right)}^2$