How to solve #int_(-pi/3)^(pi/3)# #(|x| + tan x)^2# dx ?

#int_(-pi/3)^(pi/3)# #(|x| + tan x)^2# dx

|x| is an absolute x

2 Answers
Ratnaker Mehta
Feb 20, 2018

# 2/81(81sqrt3+pi^3-27pi)#.

Explanation:

We are familiar with the following Rule :

#"For "c in [a,b] (altb), int_a^bf(x)dx=int_a^cf(x)dx+int_c^bf(x)dx#.

#:. I=int_(-pi/3)^(pi/3)f(x)dx=int_(-pi/3)^0f(x)dx+int_0^(pi/3)f(x)dx#,

#=I_1+I_2, f(x)=(|x|+tanx)^2," where, "#

#I_1=int_(-pi/3)^0f(x)dx, and, I_2=int_0^(pi/3)f(x)dx#.

#"Now, in "I_1=int_(-pi/3)^0f(x)dx, -pi/3 lt x lt 0, :. |x|=-x#.

#:. I_1=int_(-pi/3)^0(tanx-x)^2dx#.

If we subst. #x=-y," then, "dx=-dy, and,#

#"as, "x=-pi/3, y=pi/3;" whereas, "x=0, y=0#.

#:. I_1=int_(pi/3)^0[tan(-y)-(-y)]^2(-dy)#,

#=-int_(pi/3)^0(y-tany)^2dy=+int_0^(pi/3)(tany-y)^2dy#.

Because, the definite integrals are independent of the

variable, we have,

# I_1=int_0^(pi/3)(tanx-x)^2dx#.

On the similar arguments, we have,

#I_2=int_0^(pi/3)(tanx+x)^2dx#.

Thus, altogether we have,

#I=int_0^(pi/3){(tanx-x)^2+(tanx+x)^2}dx#,

#=2int_0^(pi/3)(tan^2x+x^2)dx#,

#=2int_0^(pi/3)(sec^2x-1+x^2)dx#,

#=2[tanx-x+x^3/3]_0^(pi/3)#,

#=2[{tan(pi/3)-pi/3+1/3(pi/3)^3}-0]#,

#=2(sqrt3-pi/3+pi^3/81)#.

#rArr I=2/81(81sqrt3+pi^3-27pi)#.

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