IIT JEE , AIEEE Forum - Mathematics , Integral Calculus - pls tell me the answers (salutes awaiting)

ramkumar_november

I = $\int_0^\pi\;\frac{xtanx}{secx\;+\;tanx}\;dx$ ....................(1)

apply property 4..... we get

I = $\int_0^\pi\;\frac{(\pi-x)tan(\pi-x)}{sec(\pi-x)\;+\;tan(\pi-x)}\;dx$

I = $\int_0^\pi\;\frac{(\pi-x)tanx}{secx\;+\;tanx}\;dx$ ..............(2)

adding (1) and (2) ........ we get ........

2 I = $\int_0^\pi\;\frac{\pi\;tanx}{secx\;+\;tanx}\;dx$

I = $\frac{\pi}{2}$ $\int_0^{\pi}\;tanx(secx\;-\;tanx)\;dx$

I = $\frac{\pi}{2}$ $\int_0^{\pi}\;tanxsecxdx$ - $\frac{\pi}{2}$ $\int_0^{\pi}(sec^2x\;-\;1)\;dx$

I = $\frac{\pi}{2}$ $[secx]\limits_0^\pi$ - $\frac{\pi}{2}$ ( $[tanx]_0^{\pi}$ - $\pi$ )

I = $\frac{\pi}{2}$ [ -1 -1 + $\pi$ ]

so the answer is

$\int_0^\pi\;\frac{xtanx}{secx\;+\;tanx}\;dx$ = $\frac{\pi(\pi-2)}{2}$