Question Number 100189 by Dwaipayan Shikari last updated on 25/Jun/20

∫tan^i xdx

Answered by maths mind last updated on 25/Jun/20

=∫(u^i /((1+u2)))du =∫Σ(−1)^k u^(2k+i) du =Σ(((−1)^k u^(2k+i+1) )/(2k+i+1))du =u^(i+1) Σ(((−u^2 )^k )/(k!)).((k!)/(2(((i+1)/2)+k)))+c =u^(i+1) ((1/(i+1))+Σ_(k≥1) ((k!.)/(2(k+((i+1)/2)))).(((−u^2 )^k )/(k!)))+c =(u^(i+1) /(i+1))(1+Σ_(k≥1) ((k!.(((i+1)/2)))/((k+((i+1)/2)))).(((−u^2 )^k )/(k!)))+c =(u^(i+1) /(i+1))(1+Σ_(k≥1) (((1)_k .(((i+1)/2))_k )/((((i+3)/2))_k )).(((−u^2 )^k )/(k!)))+c =(u^(i+1) /(i+1)) _2 F_1 (1,((i+1)/2);((i+3)/2);−u^2 )+c u=tan(x) 2F_1 (a,b;c;x) hyper geometric function (a_n )=Π_(k=0) ^(n−1) (a+k),