Evaluate the integral ∫0π/211+tanα⁡xdx

deiteresfp

deiteresfp

Answered

2021-12-30

Evaluate the integral
0π/211+tanαxdx

Answer & Explanation

MoxboasteBots5h

MoxboasteBots5h

Expert

2021-12-31Added 35 answers

The integral can be solved by using the fact that tan(π2x)=1tanx
abf(x)dx=abf(a+bx)dx
Let
I(α)=0π2dx1+tanαx
thenI(α)=0π2dx1+tanα(π2+0x)
=0π2dx1+1tanαx
=0π2tanαx1+tanαxdx
Adding the two I(α)s yields
2I(α)=0π211+tanαxdx+0π2tanαx1+tanαxdx
=0π2dx
=π2
I(α)=π4
Mason Hall

Mason Hall

Expert

2022-01-01Added 36 answers

Set I(α)=0π2f(α,x)dx where f(α,x):R×(0,π2)R is defined as f(α,x)=11+tanα(x)
Since both ddαf(α,x)=tanα(x)log(tan(x))(1+tanα(x))2 and f(α,x) are continuous on (0,π2)R we can apply the Leibniz integral rule obtaining
I(α)=ddα0π2f(α,x)dx=0π2ddαf(α,x)dx=0
since ddαf(α,π4+x)=ddαf(α,π4x) for x[0,π4).
Jence I(α) is constant on R and
I(α)=I(0)=0π2dx1+1=π4

Vasquez

Vasquez

Expert

2022-01-09Added 457 answers

Well, this problem has a nice symmetry. The integral can be rewritten as,
I=0π/2cosα(x)sinα(x)+cosα(x)dx
By the property of definite integrals this Integral is same as,
0π/2cosα(π2x)sinα(π2x)+cosα(π2x)dx
0π/2sinα(x)sinα(x)+cosα(x)
adding to equation we get
2I=0π/2dx
2I=π2
I=π4

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