I want to solve these Integrals 1. ∫0π211+tan2xdx 2. ∫0π21(2cos2x+sin2x)2

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I want to solve these Integrals  1. ∫0π211+tan2xdx  2. ∫0π21(2cos2x+sin2x)2

macalpinee3 · Accepted Answer

For the second integral: Substitute t=tan⁡x in order to get: ∫0π21(2cos2x+sin2x)2dx=∫0∞t2+1(2+t2)2 Observe that d(−tt2+2=t2−2(t2+2)2) Now write t2+1 as the following sum: a(t2+2)+b(t2−2) It easy to calculate that a=2+122 and b=2−122, from the system of equations a+b=1 a−b=12 Using the previous, we get ∫0∞t2+1(2+t2)2dt=a∫0∞12+t2dt+b∫0∞t2−2(2+t2)2 ∫0∞t2+1(2+t2)2dt=aarctan⁡t2424|0∞−btt2+2|0∞ ∫0∞t2+1(2+t2)2dt=aπ224 ∫0∞t2+1(2+t2dt=2+142π24

Timothy Wolff · Accepted Answer

Let  I=∫0π211+tan2xdx  Using  ∫0af(x)dx=∫0af(a−x)dx  So  I=∫0π211+cot2xdx=∫0π2tan2x1+tan2xdx  Now Adding  So we get  2I=∫0π21⋅dx=π2⇒I=π4

karton · Accepted Answer

Set I(a,b,n)=∫0π21(acos2⁡x+bsin2⁡x)ndxdI(a,b,n)da=−n∫0π2cos2⁡x(acos2⁡x+bsin2⁡x)n+1dI(a,b,n)db=−n∫0π2sin2⁡x(acos2⁡x+bsin2⁡x)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)=∫0π21acos2⁡x+bsin2⁡xdx=∫0/pi21+tan2⁡xa+btan2⁡xdxSet u=tan⁡x, thusI(a,b,1)=∫0∞1a+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,finally∫0π21(2cos2⁡x+sin2⁡x)2=π(1+2)484

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}

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