Let . If I (0) = 3, then is equal to

Engineering

Mathematics

Integration by Substitution

Question

Let $I (x) = \int \frac{6}{\sin^{2} x (1 - cotx)^{2 ∣}} dx$ . If I (0) = 3, then $I (\frac{π}{12})$ is equal to

$2 \sqrt{3}$

$3 \sqrt{3}$

$\sqrt{3}$

$6 \sqrt{3}$

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$I (x) = \int \frac{6 {cosec}^{2} x}{(1 - cotx)^{2}} dx, \frac{d}{dx} (1 - cotx) = {cosec}^{2} x$

Put t = 1 – cot x

$I (x) = \int \frac{6 dt}{t^{2}} = \frac{- 6 sinx}{(sinx - cosx)} + C$

x = 0 ⇒ 3 = 0 + C

$x = \frac{π}{12} \Rightarrow I (\frac{π}{12}) = \frac{- 6 (\frac{\sqrt{3} - 1}{2 \sqrt{2}})}{((\frac{\sqrt{3} - 1}{2 \sqrt{2}}) - (\frac{\sqrt{3} + 1}{2 \sqrt{2}}))} + 3 = 3 \sqrt{3}$