1-find-0-pi-4-dx-2-a-sinx-areal-2-c-explicite-0-pi-4-sinx-2-asinx-2-dx-

Question Number 83206 by mathmax by abdo last updated on 28/Feb/20

1) f i n d \int_{0}^{\frac{π}{4}} \frac{d x}{2 + a s i n x} (a r e a l)

2) c e x p l i c i t e \int_{0}^{\frac{π}{4}} \frac{s i n x}{{(2 + a s i n x)}^{2}} d x

Commented by mathmax by abdo last updated on 29/Feb/20

1) l e t f (a) = \int_{0}^{\frac{π}{4}} \frac{d x}{2 + a s i n x} c h a n g e m e n t t a n (\frac{x}{2}) = t g i v e

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

= \int_{0}^{\sqrt{2} - 1} \frac{d t}{t^{2} + a t + 1}

t^{2} + a t + 1 = 0 \to Δ = a^{2} - 4

i f ∣ a ∣> 2 \Rightarrow Δ > 0 \Rightarrow t_{1} = \frac{- a + \sqrt{a^{2} - 4}}{2} a n d t_{2} = \frac{- a - \sqrt{a^{2} - 4}}{2}

\Rightarrow f (a) = \int_{0}^{\sqrt{2} - 1} \frac{d t}{(t - t_{1}) (t - t_{2})} = \frac{1}{t - t_{2}} \int_{0}^{\sqrt{2} - 1} (\frac{1}{t - t_{1}} - \frac{1}{t - t_{2}}) d t

= \frac{1}{\sqrt{a^{2} - 4}} {[l n ∣ \frac{t - t_{1}}{t - t_{2}} ∣]}_{0}^{\sqrt{2} - 1} = \frac{1}{\sqrt{a^{2} - 4}} {l n ∣ \frac{\sqrt{2} - 1 - \frac{- a + \sqrt{a^{2} - 4}}{2}}{\sqrt{2} - 1 - \frac{- a - \sqrt{a^{2} - 4}}{2}} ∣

- l n ∣ \frac{- a + \sqrt{a^{2} - 4}}{- a - \sqrt{a^{2} - 4}} ∣} = \frac{1}{\sqrt{a^{2} - 4}} {l n ∣ \frac{2 \sqrt{2} - 2 + a - \sqrt{a^{2} - 4}}{2 \sqrt{2} - 2 + a + \sqrt{a^{2} - 4}} ∣

- l n ∣ \frac{a - \sqrt{a^{2} - 4}}{a + \sqrt{a^{2} - 4}} ∣

i f ∣ a ∣< 2 \Rightarrow Δ < 0 \Rightarrow f (a) = \int_{0}^{\sqrt{2} - 1} \frac{d t}{t^{2} + \frac{2 a t}{2} + \frac{a^{2}}{4} + 1 - \frac{a^{2}}{4}}

= \int_{0}^{\sqrt{2} - 1} \frac{d t}{{(t + \frac{a}{2})}^{2} + \frac{4 - a^{2}}{4}} =_{t + \frac{a}{2} = \frac{\sqrt{4 - a^{2}}}{2} u \to u = \frac{2 t + a}{\sqrt{4 - a^{2}}}} \frac{4}{4 - a^{2}} \int_{\frac{a}{\sqrt{4 - a^{2}}}}^{\frac{2 \sqrt{2} - 2 + a}{\sqrt{4 - a^{2}}}} \frac{1}{1 + u^{2}} \times \frac{\sqrt{4 - a^{2}}}{2} d u

= \frac{2}{\sqrt{4 - a^{2}}} {[a r c t a n u]}_{\frac{a}{\sqrt{4 - a^{2}}}}^{\frac{2 \sqrt{2} - 2 + a}{\sqrt{4 - a^{2}}}} = \frac{2}{\sqrt{4 - a^{2}}} {a r c t a n (\frac{2 \sqrt{2} - 2 + a}{\sqrt{4 - a^{2}}})

- a r c t a n (\frac{a}{\sqrt{4 - a^{2}}})}

Commented by mathmax by abdo last updated on 29/Feb/20

2) w e h a v e f (a) = \int_{0}^{\frac{π}{4}} \frac{d x}{2 + a s i n x} \Rightarrow f^{^{'}} (a) = - \int_{0}^{\frac{π}{4}} \frac{s i n x d x}{{(2 + a s i n x)}^{2}} \Rightarrow

\int_{0}^{\frac{π}{4}} \frac{s i n x d x}{{(2 + a s i n x)}^{2}} = - f^{^{'}} (a) r e s t t h e c a l c u l u s o f f^{^{'}} (a) \dots b e c o n t i n u e d \dots