Question and Answers Forum

All Questions      Topic List

Integration Questions

Previous in All Question      Next in All Question      

Previous in Integration      Next in Integration      

Question Number 84163 by john santu last updated on 10/Mar/20

$$\int \sin \left(50x\right)\sin {}^{49}\left(x\right)dx?$$

Answered by jagoll last updated on 10/Mar/20

$$\sin 50\mathrm{x}=\sin 49\mathrm{x}\cos \mathrm{x}+\cos 49\mathrm{x}\sin \mathrm{x}$$$$\int \sin {}^{49}\mathrm{x}\sin 49\mathrm{x}\cos \mathrm{x}\mathrm{dx}+$$$$\int \cos 49\mathrm{x}{\sin }^{50}\mathrm{x}\mathrm{dx}$$$$\mathrm{let}\mathrm{u}=\sin {}^{50}\mathrm{x}\Rightarrow \mathrm{du}=50\sin {}^{49}\mathrm{x}\cos \mathrm{x}\mathrm{dx}$$$$\mathrm{v}=\sin 49\mathrm{x}\Rightarrow \mathrm{dv}=49\cos 49\mathrm{x}\mathrm{dx}$$$$\Rightarrow \sin {}^{49}\mathrm{x}\cos \mathrm{x}\sin 49\mathrm{x}=\frac{1}{50}\mathrm{v}\frac{\mathrm{du}}{\mathrm{dx}};$$$$\Rightarrow \sin {}^{50}\mathrm{x}\cos 49\mathrm{x}=\frac{1}{50}\mathrm{u}\frac{\mathrm{dv}}{\mathrm{dx}}$$$$\mathrm{then}$$$$\int (\frac{1}{50}\mathrm{v}\frac{\mathrm{du}}{\mathrm{dx}}+\frac{1}{50}\mathrm{u}\frac{\mathrm{dv}}{\mathrm{dx}})\mathrm{dx}=$$$$\int \frac{1}{50}\frac{\mathrm{d}(\mathrm{u}.\mathrm{v})}{\mathrm{dx}}\times \mathrm{dx}=\frac{1}{50}\mathrm{u}.\mathrm{v}+\mathrm{c}$$$$=\frac{1}{50}\sin \left(49\mathrm{x}\right)\sin {}^{50}\left(\mathrm{x}\right)+\mathrm{c}$$$$$$

Commented by john santu last updated on 10/Mar/20

$$goodsir$$

Commented by TANMAY PANACEA last updated on 10/Mar/20

$$plscheck...\frac{dv}{dx}=49cos49xso\mathit{c}\mathit{o}\mathit{s}49\mathit{x}=\frac{1}{49}\frac{\mathit{d}\mathit{v}}{\mathit{d}\mathit{x}}$$$$\mathit{s}\mathit{o}{\mathit{s}\mathit{i}\mathit{n}}^{50}\mathit{x}\mathit{c}\mathit{o}\mathit{s}49\mathit{x}=\mathit{u}\times \frac{1}{49}\frac{\mathit{d}\mathit{v}}{\mathit{d}\mathit{x}}\ne \frac{1}{50}u\frac{dv}{dx}$$$$plscheck$$

Answered by TANMAY PANACEA last updated on 10/Mar/20

$$\int (sin49x.cosx.{sin}^{49}x+co49{xsin}^{50}x)dx$$$$I_1=\int sin49x.cosx.{sin}^{49}xdx$$$$I_2=\int cos49{xsin}^{50}xdx$$$$I=I_1+I_2$$$$I_1=sin49x\int {sin}^{49}xcosxdx-\int [\frac{dsin49x}{dx}\int {sin}^{49}xcosxdx]dx$$$$=sin49x.\frac{{sin}^{50}x}{50}-\int 49cos49x.\frac{{sin}^{50}x}{50}dx$$$$=\frac{sin49x.{sin}^{50}x}{50}-\frac{49}{50}{I}_2$$$$I=I_1+I_2$$$$=\frac{sin49x.{sin}^{50}x}{50}-\frac{49}{50}{I}_2+I_2$$$$I=\frac{sin49x.{sin}^{50}x}{50}+\frac{I_2}{50}$$$$waitpls$$$$$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com