Solve this: Q . P e o v e t h a t ∫ 1 e tan α 1 1 + t 2 d t + ∫ 1 e c o t α d t t 1 + t 2 = 1 , f o r a l l α f o r w h i c h tan α α a n d c o t α a r e d e f i n e d Share with your friends Share 0 Aarushi Mishra answered this Dear studentThere is a mistake in the question. It should be∫1etan αt1+t2dt+∫1ecot α1t1+t2dtLet t2=kdk=2t dt12dk=t dtI=12∫1e2tan2 α11+kdk+12∫1e2cot2 α1k1+kdk=12ln 1+k1e2tan2 α+12∫1e2cot2 α1+k-kk1+kdk=12ln 1+k1e2tan2 α+12∫1e2cot2 α1+kk1+kdk-12∫1e2cot2 αkk1+kdk=12ln 1+k1e2tan2 α+12∫1e2cot2 α1kdk-12∫1e2cot2 α11+kdk=12ln 1+k1e2tan2 α+12ln k1e2cot2 α-12ln 1+k1e2cot2 α=12ln 1+tan2 α-ln 1+1e2+12ln cot2 α-ln 1e2-12ln 1+cot2 α-ln 1+1e2We know1+cot2 α=cosec2α1+tan2 α=sec2α=12ln sec2α-ln 1+1e2+12ln cot2 α-ln 1e2-12ln cosec2α-ln 1+1e2=122 ln sec α-ln 1+1e2+122ln cot α-ln 1e2-122ln cosec α-ln 1+1e2= ln sec α-12ln 1+1e2+ln cot α-12ln 1e2-ln cosec α+12ln 1+1e2=ln sec α-ln cosec α+ln cot α-12ln 1e2=ln sec α-ln cosec α+ln cot α+12ln e2=ln sec αcosec α+ln cot α+12ln e2=ln 1cos α1sin α+ln cot α+12ln e2=ln tan α+ln cot α+ln e=ln tan α cot α+ln e=ln 1+ln e=0+1=1 0 View Full Answer