integration of cosx cos2x cos3x
Answer (1)
7th Jun, 2020
Hello there,
Let's solve this problem.
I= (int.)cosxcos2xcos3xdx
=(int.) 1/2 (2cosxcos3x)cos2xdx
=(int.) 1/2 (cos4x+cos2x) cos2xdx
= (int.)1/2 (cos4xcos2x+cos2x)dx
=(int.) 1/4 (2cosxcos2x+2cos2x)dx
= 1/4 (int.) (cos6x+cos2x+2cos2x)dx
= 1/4 (int.) (cos6x+cos2x+cos4x+1)dx
= 1/4 [ 1/6(sin6x) + 1/2(sin2x) + 1/4(sin4x) +x] +C
This is the final answer of the given expression. Since the integration sign is not visible, I have used (int.) in place of the integration sign.
Hope this answer helps you.
Good luck!!
Let's solve this problem.
I= (int.)cosxcos2xcos3xdx
=(int.) 1/2 (2cosxcos3x)cos2xdx
=(int.) 1/2 (cos4x+cos2x) cos2xdx
= (int.)1/2 (cos4xcos2x+cos2x)dx
=(int.) 1/4 (2cosxcos2x+2cos2x)dx
= 1/4 (int.) (cos6x+cos2x+2cos2x)dx
= 1/4 (int.) (cos6x+cos2x+cos4x+1)dx
= 1/4 [ 1/6(sin6x) + 1/2(sin2x) + 1/4(sin4x) +x] +C
This is the final answer of the given expression. Since the integration sign is not visible, I have used (int.) in place of the integration sign.
Hope this answer helps you.
Good luck!!
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