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Mathematics Question on General and Particular Solutions of a Differential Equation

The integral ∫dx(1+x)x−x2\displaystyle\int \frac{dx}{(1+ \sqrt{x}) \sqrt{x - x^2}} is equal to (where CC is a constant of integration)

A

−21+x1−x+C-2 \sqrt{\frac{1+ \sqrt{x}}{1 - \sqrt{x}}} + C

B

−21−x1+x+C-2 \sqrt{\frac{1- \sqrt{x}}{1+ \sqrt{x}}} + C

C

−1−x1+x+C- \sqrt{\frac{1- \sqrt{x}}{1+\sqrt{x}}} + C

D

21+x1−x+C2 \sqrt{\frac{1+ \sqrt{x}}{1 - \sqrt{x}}} + C

Answer

−21−x1+x+C-2 \sqrt{\frac{1- \sqrt{x}}{1+ \sqrt{x}}} + C

Explanation

Solution

I=∫dx(1+x)x−x2I =\int \frac{ dx }{(1+\sqrt{ x }) \sqrt{ x - x ^{2}}}
put x=cos⁡2θx =\cos ^{2} \theta
dx=−2cos⁡θsin⁡θdθdx =-2 \cos \theta \sin \theta d \theta
I=∫−2sin⁡θcos⁡θdθ(1+cos⁡θ)cos⁡θsin⁡θ     =−2∫dθ2cos⁡2θ/2I =\int \frac{-2 \sin \theta \cos \theta d \theta}{(1+\cos \theta) \cos \theta \sin \theta} \,\,\,\,\,=-2 \int \frac{ d \theta}{2 \cos ^{2} \theta / 2}
=−∫sec⁡2(θ2)dθ       ∴cos⁡θ=x=-\int \sec ^{2}\left(\frac{\theta}{2}\right) d \theta \,\,\,\,\,\,\, \therefore \cos \theta=\sqrt{ x }
=−2tan⁡θ/2+C      1−tan⁡2θ/21+tan⁡2θ/2=x=-2 \tan \theta / 2+ C \,\,\,\,\,\, \frac{1-\tan ^{2} \theta / 2}{1+\tan ^{2} \theta / 2}=\sqrt{ x }
=−21−x1+x+c=-2 \sqrt{\frac{1-\sqrt{ x }}{1+\sqrt{ x }}}+ c
⇒tan⁡2(θ2)=1−x1+x\Rightarrow \tan ^{2}\left(\frac{\theta}{2}\right)=\frac{1-\sqrt{ x }}{1+\sqrt{ x }}