Integrate #int_0^(pi/4)(1+tant)^3sec^2t# #dt# ?

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Answer 1 · 2018-06-22T04:21:50

#=int_0^(pi/4) (1+tant)^3sec^2tdt=15/4#.

Here,

#I=int_0^(pi/4) (1+tant)^3sec^2tdt#

Subst. #1+tant=u=>sec^2tdt=du#

#=>t=0 tou=1 and t=pi/4tou=2#

So,

#I=int_1^2u^3du#

#=[u^4/4]_1^2#

#=2^4/4-1^4/4#

#=16/4-1/4#

#=15/4#.

Answer 2 · 2018-06-22T04:24:36

# 15/4=3.75#.

Let, #I=int_0^(pi/4)(1+tant)^3sec^2tdt#.

Subst. #(1+tant)=x," so that, "(0+sec^2t)dt=dx#.

#"Also : "t=0, x=1+tan0=1+0=1, &, t=pi/4, x=2#.

#:. I=int_1^2x^3dx#,

#=[x^4/4]_1^2#,

#=16/4-1/4#.

# rArr I=15/4=3.75#.