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Question Number 20387 by tammi last updated on 26/Aug/17

$$\int \sin px\cos qxdx$$

Answered by mrW1 last updated on 26/Aug/17

$$\sin (\mathrm{p}+\mathrm{q})\mathrm{x}=\sin \mathrm{px}\cos \mathrm{qx}+\cos \mathrm{px}\sin \mathrm{qx}$$$$\sin (\mathrm{p}-\mathrm{q})\mathrm{x}=\sin \mathrm{px}\cos \mathrm{qx}-\cos \mathrm{px}\sin \mathrm{qx}$$$$\Rightarrow \sin \mathrm{px}\cos \mathrm{qx}=\frac{1}{2}[\sin (\mathrm{p}+\mathrm{q})\mathrm{x}+\sin (\mathrm{p}-\mathrm{q})\mathrm{x}]$$$$\Rightarrow \int \sin \mathrm{px}\cos \mathrm{qx}\mathrm{dx}=\frac{1}{2}[\int \sin (\mathrm{p}+\mathrm{q})\mathrm{x}\mathrm{dx}+\int \sin (\mathrm{p}-\mathrm{q})\mathrm{x}\mathrm{dx}]$$$$=-\frac{\cos (\mathrm{p}+\mathrm{q})\mathrm{x}}{2(\mathrm{p}+\mathrm{q})}-\frac{\cos (\mathrm{p}-\mathrm{q})\mathrm{x}}{2(\mathrm{p}-\mathrm{q})}+\mathrm{C}$$

Answered by Joel577 last updated on 26/Aug/17

$$I=\int \sin px.\cos qxdx$$$$=\frac{1}{2}\int \sin \left[(p+q)x\right]+\sin \left[(p-q)x\right]dx$$$$=\frac{1}{2}(-\frac{\cos \left[(p+q)x\right]}{p+q}-\frac{\cos \left[(p-q)x\right]}{p-q})+C$$$$=-\frac{1}{2}(\frac{\cos \left[(p+q)x\right]}{p+q}+\frac{\cos \left[(p-q)x\right]}{p-q})+C$$

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