JEE Main 2023 — Mathematics Calculus

Given,

$$\int \sqrt{\mathrm{sec}2x-1}dx=\alpha {\mathrm{log}}_{e}|\mathrm{cos}2x+\beta +\sqrt{\mathrm{cos}2x(1+\mathrm{cos}\frac{1}{\beta }x)}|+C$$∫sec2x−1​dx=αloge​∣cos2x+β+cos2x(1+cosβ1​x)​∣+C

Now solving $L.H.S$L.H.S we get,

$$\int \sqrt{\mathrm{sec}2x-1}dx=\int \sqrt{\frac{1-\mathrm{cos}2x}{\mathrm{cos}2x}}dx$$∫sec2x−1​dx=∫cos2x1−cos2x​​dx

$$\Rightarrow \int \sqrt{\mathrm{sec}2x-1}dx=\sqrt{2}\int \frac{\mathrm{sin}x}{\sqrt{2{\mathrm{cos}}^{2}x-1}}dx$$⇒∫sec2x−1​dx=2​∫2cos2x−1​sinx​dx

Now let $\mathrm{cos}x=t\Rightarrow -\mathrm{sin}xdx=dt$cosx=t⇒−sinxdx=dt

$$\Rightarrow \int \sqrt{\mathrm{sec}2x-1}dx=-\sqrt{2}\int \frac{dt}{\sqrt{2{t}^{2}-1}}$$⇒∫sec2x−1​dx=−2​∫2t2−1​dt​

$$\Rightarrow \int \sqrt{\mathrm{sec}2x-1}dx=-\mathrm{ln}|\sqrt{2}\mathrm{cos}x+\sqrt{\mathrm{cos}2x}|+C$$⇒∫sec2x−1​dx=−ln∣2​cosx+cos2x​∣+C

$$=-\frac{1}{2}\mathrm{ln}|2{\mathrm{cos}}^{2}x+\mathrm{cos}2x+2\sqrt{\mathrm{cos}2x}\cdot \sqrt{2}\mathrm{cos}x|+C$$=−21​ln∣2cos2x+cos2x+2cos2x​⋅2​cosx∣+C

$$=-\frac{1}{2}\mathrm{ln}|\mathrm{cos}2x+\frac{1}{2}+\sqrt{\mathrm{cos}2x}\cdot \sqrt{1+\mathrm{cos}2x}|+C$$=−21​ln∣cos2x+21​+cos2x​⋅1+cos2x​∣+C

Now on comparing with $\displaystyle \int \sqrt{\mathrm{sec}2x-1}dx=\alpha {\mathrm{log}}_{e}|\mathrm{cos}2x+\beta +\sqrt{\mathrm{cos}2x(1+\mathrm{cos}\frac{1}{\beta }x)}|+C$∫sec2x−1​dx=αloge​∣cos2x+β+cos2x(1+cosβ1​x)​∣+C

We get, $\beta =\frac{1}{2},\alpha =-\frac{1}{2}\Rightarrow \beta -\alpha =1$β=21​,α=−21​⇒β−α=1