Question

if $\displaystyle\sf\int\dfrac{sec^2x-2010}{sin^{2010}x}dx = \dfrac{P(x)}{sin^{2010}x}+C$ find P(π/3)​

Answer 1

we have $\displaystyle\sf\int\dfrac{sec^2x-2010}{sin^{2010}x}dx$

$\displaystyle\sf =\int sec^2 x(sin\:x)^{-2010} - 2010\int \dfrac{1}{(sin\;x)^{2010}} dx$

$\displaystyle\sf = I_1-I_2$

applying by parts on $\displaystyle\sf I_1$

$\displaystyle\sf\implies I_2 = \dfrac{tan\:x}{(sin\:x)^{2010}} + 2010\int \dfrac{tanx\:cosx}{(sinx)^{2011}} dx$

$\displaystyle\sf = \dfrac{tanx}{(sinx)^{2010}} + 2010\int \dfrac{dx}{(sinx)^{2010}}$

$\displaystyle\sf = \dfrac{tan\:x}{(sin\:x)^{2010}} = \dfrac{P(x)}{(sin\:x)^{2010}}$

$\displaystyle\sf \implies P(x) = tanx$

$\displaystyle\sf\implies P\left(\dfrac{\pi}{3}\right) = tan\left(\dfrac{\pi}{3}\right)$

$\displaystyle\sf = \sqrt{3}$