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Question Number 32156 by jarjum last updated on 20/Mar/18

$$evaluate\int \frac{\sqrt{cos2x}}{sinx}dx$$

Commented by abdo imad last updated on 20/Mar/18

$$I=\int \sqrt{\frac{cos\left(2x\right)}{{sin}^{2}x}}dx=\int \sqrt{\frac{cos\left(2x\right)}{\frac{1-cos\left(2x\right)}{2}}}dx$$$$=\int \sqrt{\frac{2cos\left(2x\right)}{1-cos\left(2x\right)}}dxch.tanx=tgive$$$$I=\int \sqrt{\frac{2\frac{1-t^2}{1+t^2}}{1-\frac{1-t^2}{1+t^2}}}\frac{dt}{1+t^2}=\int \sqrt{\frac{2(1-t^2)}{2t^2}}\frac{dt}{1+t^2}$$$$=\int \sqrt{\frac{1}{t^2}-1}\frac{dt}{1+t^2}ch.\frac{1}{t}=ugive$$$$I=\int \sqrt{u^2-1}\frac{1}{1+\frac{1}{u^2}}\frac{-du}{u^2}=-\int \frac{\sqrt{u^2-1}}{1+u^2}duch.u=ch\left(\alpha \right)$$$$I=-\int \frac{sh\left(\alpha \right)}{1+{ch}^2\left(\alpha \right)}sh\left(\alpha \right)d\alpha \Rightarrow -I=\int \frac{{sh}^2\left(\alpha \right)}{1+{ch}^2\left(\alpha \right)}d\alpha$$$$=\int \frac{\frac{1-ch\left(2\alpha \right)}{2}}{\frac{1+\frac{1+ch\left(2\alpha \right)}{2}}{}}d\alpha =\int \frac{1-ch\left(2\alpha \right)}{3+ch\left(2\alpha \right)}d\alpha$$$$=\int \frac{1-\frac{e^{2\alpha }+e^{-2\alpha }}{2}}{3+\frac{e^{2\alpha }+e^{-2\alpha }}{2}}d\alpha =\int \frac{2-e^{2\alpha }-e^{-2\alpha }}{6+e^{2\alpha }+e^{-2\alpha }}d\alpha$$$$ch.e^{2\alpha }=x\Rightarrow \alpha =\frac{1}{2}ln\left(x\right)\Rightarrow$$$$-I=\int \frac{2-x-\frac{1}{x}}{6+x+\frac{1}{x}}\frac{dx}{2x}=\int \frac{2x-x^2-1}{x(12x+2x^2+2)}dx$$$$\Rightarrow I=\int \frac{x^2-2x+1}{2x(x^2+6x+1)}dxthisintegraliscalculable$$$$afterdecompositionofthefraction....becontinued...$$

Commented by jarjum last updated on 21/Mar/18

$$thankyou$$

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