By-using-the-substitution-x-cos-2-prove-that-1-x-1-x-dx-sin-2-2-C-

Question Number 152608 by ZiYangLee last updated on 30/Aug/21

By using the substitution x = \cos 2 θ,

prove that \int \sqrt{\frac{1 + x}{1 - x}} d x = - \sin 2 θ - 2 θ + C

Answered by Olaf_Thorendsen last updated on 30/Aug/21

F (x) = \int \sqrt{\frac{1 + x}{1 - x}} d x

F (θ) = \int \sqrt{\frac{1 + \cos 2 θ}{1 - \cos 2 θ}} (- 2 \sin 2 θ d θ)

F (θ) = \int \sqrt{\frac{{2 \cos}^{2} θ}{{2 \sin}^{2} θ}} (- 4 \sin θ \cos θ d θ)

F (θ) = \int ∣ \cot θ ∣ (- 4 \sin θ \cos θ d θ)

F (θ) = - 4 \int \cos^{2} θ d θ)

F (θ) = - 2 \int (1 + \cos 2 θ) d θ)

F (θ) = - 2 (θ + \frac{1}{2} \sin 2 θ)

F (θ) = - \sin 2 θ - 2 θ + C

Commented by puissant last updated on 30/Aug/21

G e n i a l M r!!!

Answered by Ar Brandon last updated on 30/Aug/21

I = \int \sqrt{\frac{1 + x}{1 - x}} d x, x = \cos 2 ϑ

= - 2 \int \sqrt{\frac{1 + \cos 2 ϑ}{1 - \cos 2 ϑ}} \cdot \sin 2 ϑ d ϑ

= - 2 \int \frac{1 + \cos 2 ϑ}{\sqrt{1 - \cos^{2} 2 ϑ}} \cdot \sin 2 ϑ d ϑ

= - 2 \int \frac{1 + \cos 2 ϑ}{∣ \sin 2 ϑ ∣} \cdot \sin 2 ϑ d ϑ

= \mp 2 \int (1 + \cos 2 ϑ) d ϑ

= \mp (2 ϑ + \sin 2 ϑ) + C