Let I(x)=int frac 6 sin 2 x(1-cot x) 2 d x . If I(0)=3 , then I (π/12 ) is equal to :

Mathematics · Indefinite Integration

JEE Main 2024 — 8 April, Shift 1 — Question 12

Let $I(x)=\int \frac{6}{\sin ^{2} x(1-\cot x)^{2}} d x$ . If $I(0)=3$ , then $\mathrm{I}\left(\frac{\pi}{12}\right)$ is equal to :

$I(x)=\int \frac{6 d x}{\sin ^{2} x(1-\cot x)^{2}}=\int \frac{6 \operatorname{cosec}^{2} x d x}{(1-\cot x)^{2}}$

Put $1-\cot x=t$

$\operatorname{cosec}^{2} x d x=d t$

$I=\int \frac{6 \mathrm{dt}}{\mathrm{t}^{2}}=\frac{-6}{\mathrm{t}}+\mathrm{c}$

$I(x)=\frac{-6}{1-\cot x} c, c=3$

$I(x)=3-\frac{6}{1-\cot x},$

$I\left(\frac{\pi}{12}\right)=3-\frac{6}{1-(2+\sqrt{3})}$

$I\left(\frac{\pi}{12}\right)=3+\frac{6}{\sqrt{3}+1}=3+\frac{6(\sqrt{3}-1)}{2}=3 \sqrt{3}$

Answer key and solution verified before publishing.

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