Mathematics

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


SOLUTION
$$ \displaystyle I = \int \dfrac{\sqrt{cos2x}}{sinx} = \frac{cos2x}{sinx \sqrt{cos 2x}}dx $$
$$\displaystyle  = \int \dfrac{1-2sin^{2}x}{sinx\sqrt{cos2x}}dx$$
$$ \displaystyle = \int \dfrac{dx}{sinx\sqrt{cox2x}}-2\int \frac{sin2x}{sinx\sqrt{cos2x}}dx$$
$$ \displaystyle = \int \frac{dx}{sinx \sqrt{cos^{2}x-sin^{2}x}}-2\int \frac{sinx}{cos2x}dx$$
$$\displaystyle  = \int \frac{1}{sin^{2}x} \frac{1}{\sqrt{cos^{2}x-1}}dx - 2\int \dfrac{sinx\, dx}{\sqrt{cos2x}}$$
$$ I_{1}$$    $$ I_{2}$$
$$ \displaystyle I_{1} = \int \frac{cos^{2}x}{\sqrt{6t^{2}x-1}}dx$$
let t= 6+x
$$dt = -cos^{2}xdx$$
$$ \displaystyle = -\int \frac{dt}{\sqrt{t^{2}-1}} = -ln(1+\sqrt{t^{2}-1})+c_{1}$$
$$ \displaystyle I_{1} = -ln (1+\sqrt{cot^{2}x-1})+c_{1}...(1)$$
$$\displaystyle  I_{2} = 2\int \dfrac{sinx \, dx}{cos2x} = 2\int \frac{sinx}{2cos^{2}x-1}dx$$
$$ cosx = u \Rightarrow sin x dx = -du$$
$$\displaystyle \Rightarrow I_{2} =-2\int \dfrac{du}{2u^{2}-1} = \sqrt{2}\int \dfrac{du}{u^{2}-\dfrac{1}{2}} = -\sqrt{2}ln(4+\sqrt{u^{2}-\dfrac{1}{2}})+c_{2}$$
$$\displaystyle  = -\sqrt{2}ln(cosx+\sqrt{cos^{2}x-\dfrac{1}{2}})+c_{2}...(2)$$
$$ I =  I_{1}-I_{2}$$
$$ \displaystyle I= ln (1+\sqrt{cot^{2}x-1})+\sqrt{2}ln(cosx+\sqrt{cos^{2}x-\dfrac{1}{2}})+c$$
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Subjective Medium Published on 17th 09, 2020
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