Explained: "Area bounded by y=1+sin⁡{x}cos⁡{x} and y=1−sin⁡{x}cos⁡{x}"

Addison Gross

Answered question

2022-01-27

Area bounded by

y = \sqrt{\frac{1 + \sin {x}}{\cos {x}}}

and

y = \sqrt{\frac{1 - \sin {x}}{\cos {x}}}

Wilson Mitchell

Beginner2022-01-28Added 8 answers

HINT
We could use Tangent half-angle substitution to obtain

\sin x = \frac{2 t}{1 + t^{2}} \cos x = \frac{1 - t^{2}}{1 + t^{2}} t = \tan \frac{x}{2} \Rightarrow d t = \frac{12}{1 + t^{2}} d x

then

\int_{0}^{\frac{π}{4}} \sqrt{\frac{1 + \sin {x}}{\cos {x}}} - \sqrt{\frac{1 - \sin {x}}{\cos {x}}} d x = \int_{0}^{\sqrt{2} - 1} (\sqrt{\frac{{(1 + t)}^{2}}{1 - t^{2}}} - \sqrt{\frac{{(1 - t)}^{2}}{1 - t^{2}}}) \frac{2}{1 + t^{2}} d t = \dots

liep3p

Beginner2022-01-29Added 4 answers

I = \int_{0}^{\frac{π}{4}} \sqrt{\frac{1 + \sin (x)}{\cos (x)}} - \sqrt{\frac{1 - \sin (x)}{\cos (x)}} d x

using Weierstrass substitution:
we know that

\sin (x) = \frac{2 t}{1 + t^{2}}, \cos (x) = \frac{1 - t^{2}}{1 + t^{2}}, d x = \frac{2 d t}{1 + t^{2}}

so we can obtain:

I = \int_{0}^{\sqrt{2} - 1} (\sqrt{\frac{1 + t^{2} + 2 t}{1 - t^{2}}} - \sqrt{\frac{1 + t^{2} - 2 t}{1 - t^{2}}}) \cdot \frac{2 d t}{1 + t^{2}}

= \int_{0}^{\sqrt{2} - 1} (\sqrt{\frac{{(1 + t)}^{2}}{(1 - t) (1 + t)}} - \sqrt{\frac{{(1 - t)}^{2}}{(1 - t) (1 + t)}}) \cdot \frac{2 d t}{1 + t^{2}}

= \int_{0}^{\sqrt{2} - 1} (\sqrt{\frac{1 + t}{1 - t}} - \sqrt{\frac{1 - t}{1 + t}}) \cdot \frac{2 d t}{1 + t^{2}}

= \int_{0}^{\sqrt{2} - 1} \frac{(1 + t) - (1 - t)}{\sqrt{1 - t^{2}}} \cdot \frac{2 d t}{1 + t^{2}}

= \int_{0}^{\sqrt{2} - 1} \frac{4 t}{(1 + t^{2}) \sqrt{1 - t^{2}}} d t

so it is (b)

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