Question Number 105781 by IE last updated on 31/Jul/20

Commented by prakash jain last updated on 31/Jul/20

Answered by prakash jain last updated on 31/Jul/20
![∫_0 ^(π/4) tan^n xdx=∫_0 ^(π/4) tan^(n−2) x(sec^2 x−1)dx =∫_0 ^(π/4) tan^(n−2) xsec^2 xdx−∫_0 ^(π/4) tan^(n−2) xdx I_n =[((tan^(n−1) x)/(n−1))]_0 ^(π/4) −I_(n−2) I_n =(1/(n−1))−I_(n−2) I_0 =∫_0 ^(π/4) dx=(π/4) I_1 =∫_0 ^(π/4) tan xdx=((ln 2)/2) I_2 =1−I_0 I_3 =(1/2)−I_1 I_4 =(1/3)−1+I_0 I_6 =(1/5)−(1/3)+1−I_0 I_(2n) =(−1)^n [(π/4)+Σ_(m=1) ^n (((−1)^m )/(2m−1))] Question 2088 for n odd (2n+1)](https://www.tinkutara.com/question/Q105788.png)
Commented by IE last updated on 31/Jul/20

Commented by prakash jain last updated on 31/Jul/20

Commented by prakash jain last updated on 31/Jul/20

Commented by IE last updated on 31/Jul/20

Commented by IE last updated on 31/Jul/20

Commented by 1549442205PVT last updated on 01/Aug/20
