What is #int 289cos^4(17x)dx#?

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Answer 1 · 2016-07-15T12:33:30

#(17/32)(204 x + 8 sin 34 x-+sin 68 x) + C#

Use # cos^2 a = (1 + cos 2a )/2#

#289 int cos^4 (17 x) d x#

#= 289int ((1 + cos 34 x)/2)^2 d x#

#=(289/4)int ( 1 + 2 cos 34 x + co^2 34 x ) d x#

#= 289/4)int (1 + 2 cos 34 x + (1 + cos 68x)/2) d x#

#=(289/4)(x + (2/34) sin 34 x +x/2+(1/136)sin 68x) + C#

#=(17/32)(204 x + 8 sin 34 x+sin 17 x) + C#