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Question Number 193852 by SaRahAli last updated on 21/Jun/23

Answered by cortano12 last updated on 21/Jun/23

$$\frac{1}{\sqrt{2}}\int \frac{\sin 2\mathrm{x}}{\cos (\mathrm{x}-\frac{\pi }{4})}=\frac{1}{\sqrt{2}}\int \frac{\cos 2\mathrm{u}}{\cos \mathrm{u}}\mathrm{du}$$$$=\frac{1}{\sqrt{2}}\int \frac{2\cos {}^2\mathrm{u}-1}{\cos \mathrm{u}}\mathrm{du}$$$$=\frac{1}{\sqrt{2}}[\int 2\cos \mathrm{u}\mathrm{du}-\int \sec \mathrm{u}\mathrm{du}]$$$$=\frac{1}{\sqrt{2}}[2\sin \mathrm{u}-\ln \mid \sec \mathrm{u}+\tan \mathrm{u}\mid +\mathrm{c}$$$$=\frac{1}{\sqrt{2}}[2\sin (\mathrm{x}-\frac{\pi }{4})-\ln \mid \frac{1+\sin (\mathrm{x}-\frac{\pi }{4})}{\cos (\mathrm{x}-\frac{\pi }{4})}\mid ]+\mathrm{c}$$

Answered by witcher3 last updated on 21/Jun/23

$$\sin \left(2\mathrm{x}\right)=\frac{1}{2}[{(\sin \left(\mathrm{x}\right)+\cos \left(\mathrm{x}\right))}^2-{(\mathbf{sin}\left(\mathbf{x}\right)-\mathbf{cos}\left(\mathbf{x}\right))}^2]$$

Answered by MM42 last updated on 21/Jun/23

$$\int \frac{1+sin2x-1}{sinx+cosx}dx=\int \frac{{(sinx+cosx)}^2-1}{sinx+cosx}dx$$$$=\int (sinx+cosx)dx-\int \frac{dx}{\sqrt{2}sin(x+\frac{\pi }{4})}$$$$=-cosx+sinx-\frac{1}{\sqrt{2}}ln\left(tan(\frac{x}{2}+\frac{\pi }{8})\right)+c$$$$$$

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