Which is the easiest way to evaluate \int_0^{\pi/2}(\sqrt{\tam x}+\sqrt{\cot x})

osnomu3

osnomu3

Answered question

2022-01-04

Which is the easiest way to evaluate 0π2(tanx+cotx) ?

Answer & Explanation

Durst37

Durst37

Beginner2022-01-05Added 37 answers

0π2tanxdx+cotxdx
=0π2sinx+cosxsinxcosxdx=0π2sinx+cosx2sinxcosx2dx=2
0π2sinx+cosx1(sinxcosx)2dx
=20π2sinx+cosx1(sinxcosx)2dx
Let t=sinxcosx, dx=dtsinx+cosx
xπ2t=(sinxcosx)1
x0t=(sinxcosx)1
21111t2dt=2[sin1t]11=2[π2(π2)]
=2π
Mollie Nash

Mollie Nash

Beginner2022-01-06Added 33 answers

Let u=tan(x). Then u2=tan(x) and 2udu=(1+tan2(x))dx. Thus
0π2tan(x)dx=02u21+u4du
Since 1+u4=(1+2u+u2)(12u+u2), partial fraction decomposition applies:
2u21+u4=12(uu22u+1uu2+2u+1)
Hence
2u21+u4du=122log(u22u+1u2+2u+1+tan1(2+1)tan1(12u)2
Applying the fundamental theorem of calculus:
0π2tan(x)dx=π2
star233

star233

Skilled2022-01-11Added 403 answers

We will employing the substitution u=tanx:
u=1+tan2x2tanx
and
20π/2tanxdx=40u21+u4du=2u21+u4du
The last integral has two poles (u1=eiπ/4, u2=ei3π/4) in the upper complex half-plane. The corresponding residue are
Resu=u1u21+u4=u24 Resu=u1u21+u4=u14
Thus the value of the integral is
20π/2tanxdx=πi(u1+u2)=2π

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