What does cot(x−π/3)·cot(x+π/3)+1 simplify to?

It simplifies to 2/(2cos2x+1) = 2/(3−4sin²x). This is derived using product-to-sum formulas with the specific angle π/3.

What is 9α² if the integral equals α·ln(√3−1)?

9α²=12. The integral evaluates to −(2/√3)·ln(√3−1), so α=−2/√3 and 9α²=9·4/3=12.

∫_(π/6)^(π/4) [cot(x−π/3)·cot(x+π/3)+1] dx = α·ln(√3−1), find 9α²

JEERANKERS

MathIntegration

QInteger TypeDefinite Integral

If $\displaystyle\int_{\pi/6}^{\pi/4}\!\bigl[\cot\!\bigl(x-\tfrac{\pi}{3}\bigr)\cdot\cot\!\bigl(x+\tfrac{\pi}{3}\bigr)+1\bigr]\,dx=\alpha\ln_e(\sqrt{3}-1)$, then $9\alpha^2$ is equal to __________.

✅ Correct Answer

✓

Solution

Simplify cot(x−π/3)·cot(x+π/3)

Using $\cot A\cot B=\dfrac{\cos A\cos B}{\sin A\sin B}$ and product-to-sum:

$\cos(A-B)=\cos A\cos B+\sin A\sin B$
$\cos(A+B)=\cos A\cos B-\sin A\sin B$

$\therefore\quad\cos A\cos B=\dfrac{\cos(A-B)+\cos(A+B)}{2}$
$\sin A\sin B=\dfrac{\cos(A-B)-\cos(A+B)}{2}$

With $A=x-\pi/3$, $B=x+\pi/3$: $A-B=-2\pi/3$, $A+B=2x$.

$\cot A\cot B=\dfrac{\cos(-2\pi/3)+\cos(2x)}{\cos(-2\pi/3)-\cos(2x)}=\dfrac{-\frac{1}{2}+\cos 2x}{-\frac{1}{2}-\cos 2x}=\dfrac{2\cos 2x-1}{-2\cos 2x-1}$

Compute cot(x−π/3)·cot(x+π/3)+1

$\dfrac{2\cos 2x-1}{-(2\cos 2x+1)}+1=\dfrac{2\cos 2x-1-(2\cos 2x+1)}{-(2\cos 2x+1)}=\dfrac{-2}{-(2\cos 2x+1)}=\dfrac{2}{2\cos 2x+1}$

Use $2\cos 2x+1=2(1-2\sin^2 x)+1=3-4\sin^2 x$ and also note $2\cos2x+1=4\cos^2 x-1$... more simply:

$2\cos 2x+1=2(2\cos^2 x-1)+1=4\cos^2 x-1$

Also: $2\cos2x+1=1-4\sin^2 x+2=3-4\sin^2 x$

So the integrand $= \dfrac{2}{4\cos^2 x-1}=\dfrac{2}{(2\cos x-1)(2\cos x+1)}$.

Use partial fractions and integrate

$\dfrac{2}{(2\cos x-1)(2\cos x+1)}$... this approach is complex. Let's use $2\cos2x+1=3-4\sin^2x$.

Actually: $2\cos2x+1=4\cos^2x-1=(2\cos x-1)(2\cos x+1)$.

Use: $\dfrac{2}{4\cos^2x-1}$ — integrate using Weierstrass or note $4\cos^2x-1=3\cos^2x+\cos^2x-1=3\cos^2x-\sin^2x$... alternatively:

$\displaystyle\int\!\dfrac{2\,dx}{2\cos2x+1}=\int\!\dfrac{2\,dx}{3-4\sin^2x}=\int\!\dfrac{2\sec^2x\,dx}{3\sec^2x-4\tan^2x}=\int\!\dfrac{2\sec^2x\,dx}{3(1+\tan^2x)-4\tan^2x}$

$=\int\!\dfrac{2\sec^2x\,dx}{3-\tan^2x}=\int\!\dfrac{2\,dt}{3-t^2}$ where $t=\tan x$

Evaluate the integral

$\displaystyle\int_{\pi/6}^{\pi/4}\!\dfrac{2\,dt}{3-t^2}$ where $t$ goes from $\tan(\pi/6)=1/\sqrt{3}$ to $\tan(\pi/4)=1$.

$\dfrac{2}{3-t^2}=\dfrac{2}{(\sqrt{3}-t)(\sqrt{3}+t)}=\dfrac{1}{\sqrt{3}}\!\left(\dfrac{1}{\sqrt{3}-t}+\dfrac{1}{\sqrt{3}+t}\right)\cdot\dfrac{1}{\sqrt{3}}$

Actually: $\displaystyle\int\!\dfrac{2\,dt}{3-t^2}=\dfrac{1}{\sqrt{3}}\ln\!\left|\dfrac{\sqrt{3}+t}{\sqrt{3}-t}\right|+C$

At $t=1$: $\dfrac{1}{\sqrt{3}}\ln\!\left(\dfrac{\sqrt{3}+1}{\sqrt{3}-1}\right)$
At $t=1/\sqrt{3}$: $\dfrac{1}{\sqrt{3}}\ln\!\left(\dfrac{\sqrt{3}+1/\sqrt{3}}{\sqrt{3}-1/\sqrt{3}}\right)=\dfrac{1}{\sqrt{3}}\ln\!\left(\dfrac{(3+1)/\sqrt{3}}{(3-1)/\sqrt{3}}\right)=\dfrac{1}{\sqrt{3}}\ln 2$

$I=\dfrac{1}{\sqrt{3}}\!\left[\ln\!\dfrac{\sqrt{3}+1}{\sqrt{3}-1}-\ln 2\right]=\dfrac{1}{\sqrt{3}}\ln\!\dfrac{\sqrt{3}+1}{2(\sqrt{3}-1)}$

$\dfrac{\sqrt{3}+1}{2(\sqrt{3}-1)}=\dfrac{(\sqrt{3}+1)^2}{2(\sqrt{3}-1)(\sqrt{3}+1)}=\dfrac{4+2\sqrt{3}}{2\cdot2}=\dfrac{2+\sqrt{3}}{2}$

Numerically: $I\approx0.3602$ and $\alpha\ln(\sqrt{3}-1)=0.3602$, so $\alpha\approx-1.1547=-2/\sqrt{3}$.

$\alpha=-\dfrac{2}{\sqrt{3}}\quad\Rightarrow\quad 9\alpha^2=9\cdot\dfrac{4}{3}=\boxed{12}$

📘 Cotangent Product and Weierstrass Substitution

For $\cot(x-A)\cot(x+A)$: use product-to-sum formulas. With $A=\pi/3$, the expression simplifies to $\frac{2}{2\cos2x+1}=\frac{2}{3-4\sin^2x}$. Then the substitution $t=\tan x$ converts it to $\int 2dt/(3-t^2)$, a standard partial fraction integral.

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Important Concepts

Product-to-Sum for cot·cot$\cot A\cot B=\frac{\cos(A-B)+\cos(A+B)}{\cos(A-B)-\cos(A+B)}$. With $A=x-\pi/3$, $B=x+\pi/3$: $A-B=-2\pi/3$, $A+B=2x$. Then $\cos(-2\pi/3)=-1/2$.

Key Simplification$\cot(x-\pi/3)\cot(x+\pi/3)+1=\frac{2}{2\cos2x+1}$. Converting: $2\cos2x+1=3-4\sin^2x$. This is the core simplification that makes the integral tractable.

Weierstrass Substitution t=tan xDividing numerator and denominator by $\cos^2x$: $\frac{2\sec^2x}{3-\tan^2x}$. Setting $t=\tan x$, $dt=\sec^2x\,dx$ gives $\int 2dt/(3-t^2)$, a standard integral.

Standard Form ∫dt/(a²−t²)$\int\frac{dt}{a^2-t^2}=\frac{1}{2a}\ln\left|\frac{a+t}{a-t}\right|+C$. With $a=\sqrt{3}$: $\int\frac{2\,dt}{3-t^2}=\frac{1}{\sqrt{3}}\ln\left|\frac{\sqrt{3}+t}{\sqrt{3}-t}\right|+C$.

FAQs

How does cot(A)·cot(B) simplify when A+B=2x?

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Using $\cos(A\pm B)$ formulas: $\cot A\cot B=(\cos(A-B)+\cos(A+B))/(\cos(A-B)-\cos(A+B))$. Substituting $A-B=-2\pi/3$ (so $\cos=-1/2$) and $A+B=2x$ gives the simplified form.

Why substitute t=tan x?

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The denominator $3-4\sin^2x=3/\cos^2x-4\tan^2x$ after dividing by $\cos^2x$ becomes $3\sec^2x-4\tan^2x=3(1+\tan^2x)-4\tan^2x=3-\tan^2x$. So the integrand becomes $2\sec^2x/(3-\tan^2x)=2dt/(3-t^2)$.

What is ∫2dt/(3−t²)?

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$=\frac{1}{\sqrt{3}}\ln|\frac{\sqrt{3}+t}{\sqrt{3}-t}|+C$. Derived by partial fractions: $\frac{2}{3-t^2}=\frac{1}{\sqrt{3}}(\frac{1}{\sqrt{3}-t}+\frac{1}{\sqrt{3}+t})\cdot\frac{1}{\sqrt{3}}$... i.e., $\frac{1}{\sqrt{3}}(\frac{1}{\sqrt{3}+t}+\frac{1}{\sqrt{3}-t})$.

How to get α=−2/√3?

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Numerically the integral $\approx0.3602$. $\ln(\sqrt{3}-1)\approx\ln(0.732)\approx-0.3118$. So $\alpha=0.3602/(-0.3118)\approx-1.1547=-2/\sqrt{3}$.

Is this from JEE Main 2026?

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Yes, this appeared in JEE Main 2026 Section B.

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