Question

Evaluate $\int \frac{\sqrt{\cos 2x}}{\sin x}dx$

Answer 1

$I=\int \frac{\sqrt{cos2x}}{sinx}=\frac{cos2x}{sinx\sqrt{cos2x}}dx$

$=\int \frac{1-2si{n}^{2}x}{sinx\sqrt{cos2x}}dx$

$=\int \frac{dx}{sinx\sqrt{cox2x}}-2\int \frac{sin2x}{sinx\sqrt{cos2x}}dx$

$=\int \frac{dx}{sinx\sqrt{co{s}^{2}x-si{n}^{2}x}}-2\int \frac{sinx}{cos2x}dx$

$=\int \frac{1}{si{n}^{2}x}\frac{1}{\sqrt{co{s}^{2}x-1}}dx-2\int \frac{sinxdx}{\sqrt{cos2x}}$

$I_1$ $I_2$

$I_1=\int \frac{co{s}^{2}x}{\sqrt{6t^{2}x-1}}dx$

let t= 6+x

$dt=-co{s}^{2}xdx$

$=-\int \frac{dt}{\sqrt{t^2-1}}=-ln(1+\sqrt{t^2-1})+c_1$

$I_1=-ln(1+\sqrt{co{t}^{2}x-1})+c_1...\left(1\right)$

$I_2=2\int \frac{sinxdx}{cos2x}=2\int \frac{sinx}{2co{s}^{2}x-1}dx$

$cosx=u\Rightarrow sinxdx=-du$

$\Rightarrow I_2=-2\int \frac{du}{2u^2-1}=\sqrt{2}\int \frac{du}{u^2-\frac{1}{2}}=-\sqrt{2}ln(4+\sqrt{u^2-\frac{1}{2}})+c_2$

$=-\sqrt{2}ln(cosx+\sqrt{co{s}^{2}x-\frac{1}{2}})+c_2...\left(2\right)$

$I=I_1-I_2$

$I=ln(1+\sqrt{co{t}^{2}x-1})+\sqrt{2}ln(cosx+\sqrt{co{s}^{2}x-\frac{1}{2}})+c$