∫_0 ^π (x/(tan^2 (x)−1)) dx


Question and Answers Forum

All Questions Topic List
Integration Questions
Previous in All Question Next in All Question
Previous in Integration Next in Integration
Question Number 61408 by aliesam last updated on 02/Jun/19

$\int_{0}^{π} \frac{x}{{t a n}^{2} (x) - 1} d x$
	Answered by tanmay last updated on 02/Jun/19

	$\int_{0}^{π} \frac{π - x}{{t a n}^{2} (π - x) - 1} d x$ $= \int_{0}^{π} \frac{π - x}{{t a n}^{2} (x) - 1} d x$ $2 I = \int_{0}^{π} \frac{π}{{t a n}^{2} (x) - 1} d x$ $\frac{2 I}{π} = \int_{0}^{π} \frac{d x}{{t a n}^{2} (x) - 1} = \int_{0}^{π} \frac{{c o s}^{2} x}{{s i n}^{2} x - {c o s}^{2} x} d x$ $\int_{0}^{2 a} f (x) d x = 2 \int_{0}^{a} f (x) d x [w h e n f (2 a - x) = f (x)$ $\frac{2 I}{π} = 2 \int_{0}^{\frac{π}{2}} \frac{{c o s}^{2} x}{{s i n}^{2} x - {c o s}^{2} x} d x$ $\frac{I}{π} = \int_{0}^{\frac{π}{2}} \frac{{c o s}^{2} (\frac{π}{2} - x)}{{s i n}^{2} (\frac{π}{2} - x) - {c o s}^{2} (\frac{π}{2} - x)} d x$ $\frac{I}{π} = \int_{0}^{\frac{π}{2}} \frac{{s i n}^{2} x}{{c o s}^{2} x - {s i n}^{2} x} d x = \int_{0}^{\frac{π}{2}} \frac{- {s i n}^{2} x}{{s i n}^{2} x - {c o s}^{2} x} d x$ $\frac{2 I}{π} = \int_{0}^{\frac{π}{2}} \frac{{c o s}^{2} x - {s i n}^{2} x}{{s i n}^{2} x - {c o s}^{2} x} d x$ $\frac{2 I}{π} =∣ - x ∣_{0}^{\frac{π}{2}}$ $\frac{2 I}{π} = - \frac{π}{2}$ $I = \frac{- π^{2}}{4} p l s c h e c k$
	Commented by aliesam last updated on 02/Jun/19

	$t h a n k y o u s i r a b r i l l i a n t s o l$
	Commented by tanmay last updated on 02/Jun/19

	$m o s t w e l c o m e s i r . . .$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum