If In=∫cotn⁡xdx , and I0+I1+2I2+…+I8+I9+I10=Au+u22+…+u99 , where u=cot⁡x then - Sri Chaitanya Infinity Learn Best Online Courses for NCERT Solutions, CBSE, ICSE, JEE, NEET, Olympiad and Class 6 to 12

A
$A = 1$
B
$A = - 1$
C
$A = 1 / 2$
D
$A = - 1 / 2$

Solution:

$I_{n} + I_{n + 2} = \int \cot^{n} x {cosec}^{2} x d x = - \frac{1}{n + 1} \cot^{n + 1} x$

Now ,

$\begin{array}{l} I_{0} + I_{1} + 2 (I_{2} + I_{3} + \dots + I_{8}) + I_{9} + I_{10} \\ = (I_{0} + I_{2}) + (I_{1} + I_{3}) + (I_{2} + I_{4}) \\ + \dots + (I_{7} + I_{9}) + (I_{8} + I_{10}) \\ = - (u + \frac{1}{2} u^{2} + \frac{1}{3} u^{3} + \dots + \frac{1}{9} u^{9}) \end{array}$

where , $u = \cot x$

Thus , $A = - 1$

If $I_{n} = \int \cot^{n} x d x$ , and $I_{0} + I_{1} + 2 (I_{2} + \dots + I_{8})$
$+ I_{9} + I_{10} = A (u + \frac{u^{2}}{2} + \dots + \frac{u^{9}}{9})$ , where $u = \cot x$ then

Solution:

Related content