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
Questions 203525
Subjects 9
Chapters 126
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