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sin-pxcos-qxdx-

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Question Number 20387 by tammi last updated on 26/Aug/17
∫sin pxcos qxdx
∫sinpxcosqxdx
Answered by mrW1 last updated on 26/Aug/17
sin (p+q)x=sin px cos qx+cos px sin qx sin (p−q)x=sin px cos qx−cos px sin qx ⇒sin px cos qx=(1/2)[sin (p+q)x+sin (p−q)x] ⇒∫sin px cos qx dx=(1/2)[∫sin (p+q)x dx+∫sin (p−q)x dx] =−((cos (p+q)x)/(2(p+q)))−((cos (p−q)x)/(2(p−q)))+C
sin(p+q)x=sinpxcosqx+cospxsinqxsin(p−q)x=sinpxcosqx−cospxsinqx⇒sinpxcosqx=12[sin(p+q)x+sin(p−q)x]⇒∫sinpxcosqxdx=12[∫sin(p+q)xdx+∫sin(p−q)xdx]=−cos(p+q)x2(p+q)−cos(p−q)x2(p−q)+C
Answered by Joel577 last updated on 26/Aug/17
I = ∫ sin px . cos qx dx = (1/2)∫ sin [(p + q)x] + sin [(p − q)x] dx = (1/2)(−((cos [(p + q)x])/(p + q)) − ((cos [(p − q)x])/(p − q))) + C = −(1/2)(((cos [(p + q)x])/(p + q)) + ((cos [(p − q)x])/(p − q))) + C
I=∫sinpx.cosqxdx=12∫sin[(p+q)x]+sin[(p−q)x]dx=12(−cos[(p+q)x]p+q−cos[(p−q)x]p−q)+C=−12(cos[(p+q)x]p+q+cos[(p−q)x]p−q)+C

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