=1/2[sin7x sin(-x)]
=1/2[sin7x sinx]
=1/2(sin7x-sinx)dx
=1/2[-cos7x/7 -(-cos)]
I=1/2[cos-1/7cos7 c
Written on Feb. 13, 2018, 7:03 p.m.
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Answer of this question
Sin3x cos4x
= 1/2[sin(3x 4x) sin(3x-4x)]
=1/2[sin7x sin(-x)]
=1/2[sin7x sinx]
=1/2(sin7x-sinx)dx
=1/2[-cos7x/7 -(-cos)]
I=1/2[cos-1/7cos7 c
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