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0-pi-2-tan-3-xdx-




Question Number 65950 by ajfour last updated on 06/Aug/19
∫_0 ^( π/2) tan^3 xdx = ?
0π/2tan3xdx=?
Answered by Tupac Shakur last updated on 06/Aug/19
=∫_0 ^(π/2) ((sin^3 (x))/(cos^3 (x)))dx=∫_0 ^(π/2) ((sin(x){1−cos^2 (x)})/(cos^3 (x)))dx  let u=cos(x)==>du=−sin(x)dx  ∫_0 ^(π/2) ((sin(x){1−cos^2 (x)})/(cos^3 (x)))dx=∫_0 ^1 ((1−u^2 )/u^3 )du=lim_(t→0) ∫_t ^1 (1/u^3 )−(1/u^ )du=limt→0{−(3/2)+(1/(2t^2 ))+ln(t)}=+∞
=0π2sin3(x)cos3(x)dx=0π2sin(x){1cos2(x)}cos3(x)dxletu=cos(x)==>du=sin(x)dx0π2sin(x){1cos2(x)}cos3(x)dx=011u2u3du=limt0t11u31udu=limt0{32+12t2+ln(t)}=+
Answered by Tanmay chaudhury last updated on 06/Aug/19
∫tan^2 x.tanxdx  ∫(sec^2 x−1)tanxdx  ∫tanx.d(tanx)−∫tanxdx  ((tan^2 x)/2)−lnsecx+c  when we put x=(π/2) expression→∞  so some doubt ...
tan2x.tanxdx(sec2x1)tanxdxtanx.d(tanx)tanxdxtan2x2lnsecx+cwhenweputx=π2expressionsosomedoubt
Answered by ajfour last updated on 06/Aug/19
∫_0 ^( π/2)  ((tan^3 xsec^2 xdx)/((1+tan^2 x)))  let  tan x=t  ∫_0 ^( ∞) ((t^3 dt)/(1+t^2 )) = ∫_0 ^∞  ((t(t^2 +1−1)dt)/(1+t^2 ))  =∫_0 ^( c→∞) [(tdt)−(1/2)(((2tdt)/(1+t^2 )))]   = lim_(b→∞) [(b/2)−(1/2)ln (1+b)]  f(b)=lim_(b→∞) ((ln (1+b))/b)=lim_(b→∞) (1/(1+b)) →0  ⇒ Integral →∞  (but answer in book R.S. Agarwal  for class XII is a finite one !)
0π/2tan3xsec2xdx(1+tan2x)lettanx=t0t3dt1+t2=0t(t2+11)dt1+t2=0c[(tdt)12(2tdt1+t2)]=limb[b212ln(1+b)]f(b)=limbln(1+b)b=limb11+b0Integral(butanswerinbookR.S.AgarwalforclassXIIisafiniteone!)
Commented by MJS last updated on 06/Aug/19
post his answer please...
posthisanswerplease
Answered by MJS last updated on 06/Aug/19
∫tan^n  x dx=(1/(n−1))tan^(n−1)  x −∫tan^(n−2)  x dx  ∫tan^3  x dx=(1/2)tan^2  x −∫tan x dx=  =(1/2)tan^2  x +ln cos x +C  ∫_0 ^(π/2) tan^n  x dx=+∞ for n∈N^★   what′s obvious because tan^n  x ≥1 for (π/4)≤x<(π/2)  and n∈N^★
tannxdx=1n1tann1xtann2xdxtan3xdx=12tan2xtanxdx==12tan2x+lncosx+Cπ20tannxdx=+fornNwhatsobviousbecausetannx1forπ4x<π2andnN