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Question Number 58362 by peter frank last updated on 22/Apr/19
∫sinxsin3xdx
Answered by mr W last updated on 22/Apr/19
=∫sinx3sinx−4sin3xdx=∫13−4sin2xdx=∫13−2(1−cos2x)dx=∫11+2cos2xdx=14∫112+cos2xd2x=14∫112+costdt=123ln∣tant2+3tant2−3∣+C=123ln∣tanx+3tanx−3∣+C
Commented by peter frank last updated on 23/Apr/19
thanks
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