Question Number 20387 by tammi last updated on 26/Aug/17
$$\int\mathrm{sin}\:{px}\mathrm{cos}\:{qxdx} \\ $$
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](https://www.tinkutara.com/question/Q20398.png)
$$\mathrm{sin}\:\left(\mathrm{p}+\mathrm{q}\right)\mathrm{x}=\mathrm{sin}\:\mathrm{px}\:\mathrm{cos}\:\mathrm{qx}+\mathrm{cos}\:\mathrm{px}\:\mathrm{sin}\:\mathrm{qx} \\ $$$$\mathrm{sin}\:\left(\mathrm{p}−\mathrm{q}\right)\mathrm{x}=\mathrm{sin}\:\mathrm{px}\:\mathrm{cos}\:\mathrm{qx}−\mathrm{cos}\:\mathrm{px}\:\mathrm{sin}\:\mathrm{qx} \\ $$$$\Rightarrow\mathrm{sin}\:\mathrm{px}\:\mathrm{cos}\:\mathrm{qx}=\frac{\mathrm{1}}{\mathrm{2}}\left[\mathrm{sin}\:\left(\mathrm{p}+\mathrm{q}\right)\mathrm{x}+\mathrm{sin}\:\left(\mathrm{p}−\mathrm{q}\right)\mathrm{x}\right] \\ $$$$\Rightarrow\int\mathrm{sin}\:\mathrm{px}\:\mathrm{cos}\:\mathrm{qx}\:\mathrm{dx}=\frac{\mathrm{1}}{\mathrm{2}}\left[\int\mathrm{sin}\:\left(\mathrm{p}+\mathrm{q}\right)\mathrm{x}\:\mathrm{dx}+\int\mathrm{sin}\:\left(\mathrm{p}−\mathrm{q}\right)\mathrm{x}\:\mathrm{dx}\right] \\ $$$$=−\frac{\mathrm{cos}\:\left(\mathrm{p}+\mathrm{q}\right)\mathrm{x}}{\mathrm{2}\left(\mathrm{p}+\mathrm{q}\right)}−\frac{\mathrm{cos}\:\left(\mathrm{p}−\mathrm{q}\right)\mathrm{x}}{\mathrm{2}\left(\mathrm{p}−\mathrm{q}\right)}+\mathrm{C} \\ $$
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](https://www.tinkutara.com/question/Q20399.png)
$${I}\:=\:\int\:\mathrm{sin}\:{px}\:.\:\mathrm{cos}\:{qx}\:{dx} \\ $$$$\:\:\:\:=\:\frac{\mathrm{1}}{\mathrm{2}}\int\:\mathrm{sin}\:\left[\left({p}\:+\:{q}\right){x}\right]\:+\:\mathrm{sin}\:\left[\left({p}\:−\:{q}\right){x}\right]\:{dx} \\ $$$$\:\:\:\:=\:\frac{\mathrm{1}}{\mathrm{2}}\left(−\frac{\mathrm{cos}\:\left[\left({p}\:+\:{q}\right){x}\right]}{{p}\:+\:{q}}\:−\:\frac{\mathrm{cos}\:\left[\left({p}\:−\:{q}\right){x}\right]}{{p}\:−\:{q}}\right)\:+\:{C} \\ $$$$\:\:\:\:=\:−\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{cos}\:\left[\left({p}\:+\:{q}\right){x}\right]}{{p}\:+\:{q}}\:+\:\frac{\mathrm{cos}\:\left[\left({p}\:−\:{q}\right){x}\right]}{{p}\:−\:{q}}\right)\:+\:{C} \\ $$