How do you find the antiderivative of #cos^2 (x)#?

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8
Jul 30, 2016

Remember that #cos2x=cos^2x-sin^2x=>cos2x=2cos^2x-1=> cos^2x=1/2(1+cos2x)#

Hence

#int cos^2xdx=int 1/2 (1+cos2x)dx=1/2int 1dx+1/2intcos2xdx= 1/2x+1/4sin2x+c#

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Write your answer here...
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Then teach the underlying concepts
Don't copy without citing sources
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Answer

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Answer:

Explanation

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3
Jul 30, 2016

Answer:

#int cos^2(x) d x=x/2+(cos(x) sin(x))/2+C#

Explanation:

#int cos^2(x) d x=?#

#"let us use the reduction formula :"#

#cos^n(x) d x=(n-1)/(n)int cos^(n-2) (x) d x+(cos^(n-1)(x) sin (x))/n#

#"Apply n=2"#

#int cos^2(x) d x=(2-1)/2 int cos^(2-2)(x) d x+(cos^(2-1)(x) sin(x))/2#

#int cos^2(x) d x=1/2 int cos^0 (x) d x +(cos (x) sin (x))/2#

#int cos^2(x) d x=1/2 int d x+(cos (x) sin(x))/2#

#int cos^2(x) d x=1/2 x +(cos(x) sin(x))/2#

#int cos^2(x) d x=x/2+(cos(x) sin(x))/2+C#

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