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Question Number 106807 by Study last updated on 07/Aug/20

$$\int sin\left(ln3x\right)dx=???$$

Answered by bemath last updated on 08/Aug/20

$$@\mathrm{bemath}@$$$$\ln 3\mathrm{x}=\mathrm{u}\Rightarrow 3\mathrm{x}={\mathrm{e}}^{\mathrm{u}};\mathrm{dx}=\frac{{\mathrm{e}}^{\mathrm{u}}\mathrm{du}}{3}$$$$\mathrm{J}=\int \sin \left(\mathrm{u}\right)\left(\frac{{\mathrm{e}}^{\mathrm{u}}}{3}\mathrm{du}\right)$$$$\mathrm{J}=\frac{1}{3}\int {\mathrm{e}}^{\mathrm{u}}\sin \mathrm{u}\mathrm{du}\left[\mathrm{by}\mathrm{parts}\right]$$$$\mathrm{J}=\frac{1}{3}[-{\mathrm{e}}^{\mathrm{u}}\cos \mathrm{u}+\int {\mathrm{e}}^{\mathrm{u}}\cos \mathrm{u}\mathrm{du}]$$$$\mathrm{J}=-\frac{1}{3}{\mathrm{e}}^{\mathrm{u}}\cos \mathrm{u}+\frac{1}{3}[{\mathrm{e}}^{\mathrm{u}}\sin \mathrm{u}-\int {\mathrm{e}}^{\mathrm{u}}\sin \mathrm{u}\mathrm{du}]$$$$\mathrm{J}+\frac{1}{3}\mathrm{J}=-\frac{1}{3}{\mathrm{e}}^{\mathrm{u}}(\cos \mathrm{u}-\sin \mathrm{u})$$$$\mathrm{J}=\frac{3}{4}\times (-\frac{1}{3}{\mathrm{e}}^{\mathrm{u}}(\cos \mathrm{u}-\sin \mathrm{u}))+\mathrm{C}$$$$\mathrm{J}=-\frac{3\mathrm{x}}{4}(\cos \left(\ln 3\mathrm{x}\right)-\sin \left(\ln 3\mathrm{x}\right))+\mathrm{C}$$

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