JEE Main 2022 (Online) 26th July Evening ShiftQuestion 1
The integral $\int\frac{\left(1-\frac{1}{\sqrt3}\right)(\cos x-\sin x)}{\left(1+\frac{2}{\sqrt3}\sin2x\right)}dx$ is equal to
Indefinite Integration
Solution
Concept. The numerator has $\cos x-\sin x=\frac{d}{dx}(\sin x+\cos x)$ and $\sin2x=t^2-1$; the integral becomes $\int\frac{dt}{t^2-k^2}$. Then convert back to the $\tan$ form of the options. Formulas used. (1) $t=\sin x+\cos x=\sqrt2\sin\big(x+\frac\pi4\big)$: $dt=(\cos x-\sin x)dx$, $\sin2x=t^2-1$(2) $\int\dfrac{dt}{t^2-k^2}=\dfrac1{2k}\ln\Big|\dfrac{t-k}{t+k}\Big|$(3) $\dfrac{\sin A-\sin B}{\sin A+\sin B}=\dfrac{\tan\frac{A-B}2}{\tan\frac{A+B}2}$Step 1 - Denominator. By (1), $$1+\frac2{\sqrt3}(t^2-1)=\frac2{\sqrt3}\big(t^2-k^2\big),\quad k^2=\frac{2-\sqrt3}2,\ k=\frac{\sqrt3-1}2.$$ Step 2 - Integrate. $\big(1-\frac1{\sqrt3}\big)\frac{\sqrt3}2=\frac{\sqrt3-1}2=k$. By (2), $$I=k\cdot\frac1{2k}\ln\Big|\frac{t-k}{t+k}\Big|=\frac12\ln\Big|\frac{t-k}{t+k}\Big|+C.$$ Step 3 - Back to $x$. $k=\sqrt2\sin\frac\pi{12}$, so $\frac{t-k}{t+k}=\frac{\sin(x+\frac\pi4)-\sin\frac\pi{12}}{\sin(x+\frac\pi4)+\sin\frac\pi{12}}$. By (3) with $A=x+\frac\pi4$, $B=\frac\pi{12}$: $$I=\frac12\log_e\Bigg|\frac{\tan\big(\frac x2+\frac\pi{12}\big)}{\tan\big(\frac x2+\frac\pi6\big)}\Bigg|+C.$$ Answer: (A)