0-pi-x-tan-2-x-1-dx-

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 \dots

Facebook Tweet Pin