1Vaibhav ... ur absolutely right ...
I think ur stuck in the step where the expr. is ....
∫√(asecθ + atanθ) asec2θ dθ ....
Amn't I right ?????!!!!!
5 Answers
11Q→1
I = \int \sqrt{x + \sqrt{x^{2} + a^{2}}}
{x + \sqrt{x^{2}+ a^{2}}} = t → eq1
\left(1 + x/\sqrt{x^{2} + a ^{2}} \right)dx = dt
\left[ \left(\sqrt{x^{2}+ a^{2}} + x \right)/\left(\sqrt{x^{2} + a^{2}} \right)\right]dx = dt →eq3
also by rationalisation you can see that ,
-a2t = \left(x - \sqrt{x^{2} + a^{2}} \right) → eq2
subtracting eq1 - eq2 we get,
t + a2t = 2 \sqrt{x^{2} + a^{2}}
ie \sqrt{x^{2} + a^{2}} = \frac{t^{2} + a^{2}}{2t}
putting this value of \sqrt{x^{2} + a^{2}} in eq3 we get,
(\frac{2t^{2}}{a^{2} + t^{2}})dx = dt
therefore,
I = \int t^\frac{1}{2}( \frac{1}{2} + \frac{a^{2}}{2t^{2}})dt
I = \frac{t^{\frac{3}{2}}}{3} - \frac{a^{2}}{t^{\frac{1}{2}}}
(we know the value of t from eq 1 )
PLZ DO TELL ANY MISTAKES IF ANY
11i think there can be another smaller solution by putting x= atanθ
but im not able to proceed after a certain step in this method plz someone try this method also
11The other method is taking x = atanθ
Now the expr. boils down to ....
√∫(asecθ + atanθ) asec2θ dθ
=∫√a(1 + sinθcosθ) asec2θ dθ
=∫√a((sinθ/2 + cosθ/2)2cos2θ/2 - sin2θ/2) asec2dθ
=∫√a((tanθ/2 +1)21 - tan2θ/2) asec2θ dθ
Now take tanθ/2 = z
=2a√a∫√((z + 1)21 - z2) dz
=2a√a∫(z + 1) √(1 - z2) dz
=2a√a∫zdz√(1 - z2) + 2a√a∫dz√(1 - z2) ... is the simplified expr ...
Now this can be easily evaluated .... hope there r no calculation errors ...
If there r any ... please pardon me ...
1Let .............. x + √x 2 + a 2 = z 2
( x - z 2 ) 2 = x 2 + a 2 ;
or , x 2 - 2 z 2 x + z 4 = x 2+ a 2
or , 2 z 2 x = z 4 - a 2
or , x = z 22 - a 22 z 2
or , dx = dz { z2 + a 2z 3 }
Now , I = ∫ z { z2 + a 2z 3 } dz
Which is not troublesome , I think ?