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Question: If I<sub>n</sub> = \(\int_{}^{}\cot^{n}\)x dx, then I<sub>0</sub> + I<sub>1</sub> + 2(I<sub>2</sub> ...

If In = ∫cot⁡n\int_{}^{}\cot^{n}x dx, then I0 + I1 + 2(I2 + I3+ …. + I8) + I9 +

I10 =

A

–∑k=19cot⁡kxk\sum_{k = 1}^{9}\frac{\cot^{k}x}{k}

B

∑k=19cot⁡kxk!\sum_{k = 1}^{9}\frac{\cot^{k}x}{k!}

C

∑k=110cot⁡kx10\sum_{k = 1}^{10}\frac{\cot^{k}x}{10}

D

–∑k=110kcot⁡kx–\sum_{k = 1}^{10}{k\cot^{k}x}

Answer

–∑k=19cot⁡kxk\sum_{k = 1}^{9}\frac{\cot^{k}x}{k}

Explanation

Solution

In =∫cot⁡n\int_{}^{}\cot^{n}x dx =∫cot⁡n−2\int_{}^{}\cot^{n - 2}x cot2x dx

= ∫cot⁡n−2\int_{}^{}\cot^{n - 2}x(cosec2x – 1)dx = ∫cot⁡n−2\int_{}^{}\cot^{n - 2}x cosec2 x dx – In–2

Thus, In + In – 2 = – cot⁡n−1xn−1\frac{\cot^{n - 1}x}{n - 1} ...(i)

I0 + I1 + 2(I2 + …… + I8) + I9 + I10

= (I0 + I2) + (I3 + I1) + (I4 + I2) + (I5 + I3) + (I6 + I4) +(I7 + I5) + (I8 + I6) + (I9 + I7) + (I10 + I8) = – (cot⁡x1+cot⁡2x2+....+cot⁡9x9)\left( \frac{\cot x}{1} + \frac{\cot^{2}x}{2} + .... + \frac{\cot^{9}x}{9} \right) [using (i)]

= –∑k=19cot⁡kxk\sum_{k = 1}^{9}\frac{\cot^{k}x}{k}