Question-89324

Question Number 89324 by 174 last updated on 16/Apr/20

Commented by mathmax by abdo last updated on 16/Apr/20

l e t f (a) = \int_{0}^{\frac{π}{2}} l n (a + {c o s}^{2} x) d x w i t h a > 0

f^{^{'}} (a) = \int_{0}^{\frac{π}{2}} \frac{d x}{a + {c o s}^{2} x} d x = \int_{0}^{\frac{π}{2}} \frac{d x}{a + \frac{1}{1 + {t a n}^{2} x}}

= \int_{0}^{\frac{π}{2}} \frac{1 + {t a n}^{2} x}{a + {a t a n}^{2} x + 1} d x =_{t a n x = t} \int_{0}^{\infty} \frac{1 + t^{2}}{{a t}^{2} + a + 1} \frac{d t}{1 + t^{2}}

= \int_{0}^{\infty} \frac{d t}{{a t}^{2} + a + 1} = \frac{1}{a} \int_{0}^{\infty} \frac{d t}{t^{2} + \frac{a + 1}{a}} =_{t = \sqrt{\frac{a + 1}{a}} u}

\frac{1}{a} \int_{0}^{\infty} \frac{1}{\frac{a + 1}{a} (1 + u^{2})} \times \sqrt{\frac{a + 1}{a}} d u

= \frac{1}{\sqrt{a} \sqrt{a + 1}} \times \frac{π}{2} = \frac{π}{2 \sqrt{a} \sqrt{a + 1}} \Rightarrow f (a) = \frac{π}{2} \int \frac{d a}{\sqrt{a} \sqrt{a + 1}}

w e h a v e \int \frac{d a}{\sqrt{a} \sqrt{a + 1}} =_{\sqrt{a} = z} \int \frac{2 z d z}{z \sqrt{z^{2} + 1}} = 2 \int \frac{d z}{\sqrt{z^{2} + 1}}

= 2 l n (z + \sqrt{1 + z^{2}}) + c = 2 l n (\sqrt{a} + \sqrt{1 + a}) + c

\Rightarrow f (a) = π l n (\sqrt{a} + \sqrt{a + 1}) + C

{l i m}_{a \to 0^{+}} f (a) = C = 2 \int_{0}^{\frac{π}{2}} l n (c o s x) d x = 2 (- \frac{π}{2} l n (2))

= - π l n 2 \Rightarrow f (a) = π l n (\sqrt{a} + \sqrt{a + 1}) - π l n (2)

\int_{0}^{\frac{π}{2}} l n (1 + {c o s}^{2} x) d x = f (1) = π l n (1 + \sqrt{2}) - π l n (2)

= π {l n (1 + \sqrt{2}) - l n (2)}