I want to solve these Integrals 1. \int_0^{\frac{\pi}{2}}\frac{1}{1+\tan^{\sqrt{2}}x}dx 2. \int_0^{\frac{\pi}{2}}\frac{1}{(\sqrt{2}\cos^2x+\sin^2x)^2}

Ashley Bell 2022-01-01 Answered
I want to solve these Integrals
1. 0π211+tan2xdx
2. 0π21(2cos2x+sin2x)2
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Expert Answer

macalpinee3
Answered 2022-01-02 Author has 29 answers

For the second integral:
Substitute t=tanx in order to get:
0π21(2cos2x+sin2x)2dx=0t2+1(2+t2)2
Observe that
d(tt2+2=t22(t2+2)2)
Now write t2+1 as the following sum:
a(t2+2)+b(t22)
It easy to calculate that a=2+122 and b=2122, from the system of equations
a+b=1
ab=12
Using the previous, we get
0t2+1(2+t2)2dt=a012+t2dt+b0t22(2+t2)2
0t2+1(2+t2)2dt=aarctant2424|0btt2+2|0
0t2+1(2+t2)2dt=aπ224
0t2+1(2+t2dt=2+142π24

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Timothy Wolff
Answered 2022-01-03 Author has 26 answers
Let
I=0π211+tan2xdx
Using
0af(x)dx=0af(ax)dx
So
I=0π211+cot2xdx=0π2tan2x1+tan2xdx
Now Adding
So we get
2I=0π21dx=π2I=π4
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karton
Answered 2022-01-09 Author has 368 answers

Set
I(a,b,n)=0π21(acos2x+bsin2x)ndxdI(a,b,n)da=n0π2cos2x(acos2x+bsin2x)n+1dI(a,b,n)db=n0π2sin2x(acos2x+bsin2x)n+1ThereforedI(a,b,n)da+dI(a,b,n)db=nI(a,b,n+1)we haveI(a,b,n+1)=1n(dI(a,b,n)da+dI(a,b,n)db)Now we should compute I(a,b,1)I(a,b,1)=0π21acos2x+bsin2xdx=0/pi21+tan2xa+btan2xdxSet u=tanx, thusI(a,b,1)=01a+bu2dx=π2absodI(a,b,1)da=πb4aabsimilarlydI(a,b,1)db=πa4bbaapplyI(a,b,2)=(dI(a,b,1)da+dI(a,b,1)db)=π(a+b)4ababseta=2 and b=1,finally0π21(2cos2x+sin2x)2=π(1+2)484

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