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Q11  Find the integrals of the functions \cos ^ 4 2x

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 cos^42x=cos^32xcos2x

now using the identity  

cos^3x=\frac{cos3x+3cosx}{4}

cos^32xcos2x=\frac{cos6x +3cos2x}{4}cos2x=\frac{cos6xcos2x+3cos^22x}{4}

now using the below two identities 

cosacosb=\frac{cos(a+b)+cos(a-b)}{2}\\and\ cos^22x=\frac{1+cos4x}{2}\\

the value 

cos^42x=cos^32xcos2x\\=\frac{cos6xcos2x+3cos^22x}{4}=\frac{cos 4x+cos8x}{8}+\frac{3}{4}\frac{1+cos4x}{2}.

the integral of the given function can be written as

\int cos^42x=\int \frac{cos 4x+cos8x}{8}+\int \frac{3}{4}\frac{1+cos4x}{2}\\ \\=\frac{3}{8}x+\frac{sin4x}{8}+\frac{sin8x}{64}+C

 

 

Posted by

manish

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