1) find ∫_0 ^(π/4) (dx/(2+a sinx)) (areal) 2) c explicite ∫


Question and Answers Forum

All Questions Topic List
Integration Questions
Previous in All Question Next in All Question
Previous in Integration Next in Integration
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
$1) let f(a)=∫_0 ^(π/4) (dx/(2+asinx)) changement tan((x/2))=t give f(a) =∫_0 ^((√2)−1) ((2dt)/((1+t^2 )(2+a((2t)/(1+t^2 ))))) =2∫_0 ^((√2)−1) (dt/(2+2t^2 +2at)) =∫_0 ^((√2)−1) (dt/(t^2 +at +1)) t^2 +at+1=0→Δ=a^2 −4 if ∣a∣>2 ⇒ Δ>0 ⇒t_1 =((−a+(√(a^2 −4)))/2) and t_2 =((−a−(√(a^2 −4)))/2) ⇒f(a) =∫_0 ^((√2)−1) (dt/((t−t_1 )(t−t_2 ))) =(1/(t−t_2 ))∫_0 ^((√2)−1) ((1/(t−t_1 ))−(1/(t−t_2 )))dt =(1/(√(a^2 −4))) [ln∣((t−t_1 )/(t−t_2 ))∣]_0 ^((√2)−1) =(1/(√(a^2 −4))){ln∣(((√2)−1−((−a+(√(a^2 −4)))/2))/((√2)−1−((−a−(√(a^2 −4)))/2)))∣ −ln∣((−a+(√(a^2 −4)))/(−a−(√(a^2 −4))))∣}=(1/(√(a^2 −4))){ln∣((2(√2)−2+a−(√(a^2 −4)))/(2(√2)−2+a+(√(a^2 −4))))∣ −ln∣((a−(√(a^2 −4)))/(a+(√(a^2 −4))))∣ if ∣a∣<2 ⇒Δ<0 ⇒f(a)=∫_0 ^((√2)−1) (dt/(t^2 +((2at)/2) +(a^2 /4)+1−(a^2 /4))) =∫_0 ^((√2)−1) (dt/((t+(a/2))^2 +((4−a^2 )/4))) =_(t+(a/2)=((√(4−a^2 ))/2)u→u=((2t+a)/(√(4−a^2 )))) (4/(4−a^2 )) ∫_(a/(√(4−a^2 ))) ^((2(√2)−2+a)/(√(4−a^2 ))) (1/(1+u^2 ))×((√(4−a^2 ))/2)du =(2/(√(4−a^2 )))[arctanu]_(a/(√(4−a^2 ))) ^((2(√2)−2+a)/(√(4−a^2 ))) =(2/(√(4−a^2 ))){ arctan(((2(√2)−2+a)/(√(4−a^2 )))) −arctan((a/(√(4−a^2 ))))}$
$1) l e t f (a) = \int_{0}^{\frac{π}{4}} \frac{d x}{2 + a s i n x} c h a n g e m e n t t a n (\frac{x}{2}) = t g i v e$ $f (a) = \int_{0}^{\sqrt{2} - 1} \frac{2 d t}{(1 + t^{2}) (2 + a \frac{2 t}{1 + t^{2}})} = 2 \int_{0}^{\sqrt{2} - 1} \frac{d t}{2 + 2 t^{2} + 2 a t}$ $= \int_{0}^{\sqrt{2} - 1} \frac{d t}{t^{2} + a t + 1}$ $t^{2} + a t + 1 = 0 \to Δ = a^{2} - 4$ $i f ∣ a ∣> 2 \Rightarrow Δ > 0 \Rightarrow t_{1} = \frac{- a + \sqrt{a^{2} - 4}}{2} a n d t_{2} = \frac{- a - \sqrt{a^{2} - 4}}{2}$ $\Rightarrow f (a) = \int_{0}^{\sqrt{2} - 1} \frac{d t}{(t - t_{1}) (t - t_{2})} = \frac{1}{t - t_{2}} \int_{0}^{\sqrt{2} - 1} (\frac{1}{t - t_{1}} - \frac{1}{t - t_{2}}) d t$ $= \frac{1}{\sqrt{a^{2} - 4}} {[l n ∣ \frac{t - t_{1}}{t - t_{2}} ∣]}_{0}^{\sqrt{2} - 1} = \frac{1}{\sqrt{a^{2} - 4}} {l n ∣ \frac{\sqrt{2} - 1 - \frac{- a + \sqrt{a^{2} - 4}}{2}}{\sqrt{2} - 1 - \frac{- a - \sqrt{a^{2} - 4}}{2}} ∣$ $- l n ∣ \frac{- a + \sqrt{a^{2} - 4}}{- a - \sqrt{a^{2} - 4}} ∣} = \frac{1}{\sqrt{a^{2} - 4}} {l n ∣ \frac{2 \sqrt{2} - 2 + a - \sqrt{a^{2} - 4}}{2 \sqrt{2} - 2 + a + \sqrt{a^{2} - 4}} ∣$ $- l n ∣ \frac{a - \sqrt{a^{2} - 4}}{a + \sqrt{a^{2} - 4}} ∣$ $i f ∣ a ∣< 2 \Rightarrow Δ < 0 \Rightarrow f (a) = \int_{0}^{\sqrt{2} - 1} \frac{d t}{t^{2} + \frac{2 a t}{2} + \frac{a^{2}}{4} + 1 - \frac{a^{2}}{4}}$ $= \int_{0}^{\sqrt{2} - 1} \frac{d t}{{(t + \frac{a}{2})}^{2} + \frac{4 - a^{2}}{4}} =_{t + \frac{a}{2} = \frac{\sqrt{4 - a^{2}}}{2} u \to u = \frac{2 t + a}{\sqrt{4 - a^{2}}}} \frac{4}{4 - a^{2}} \int_{\frac{a}{\sqrt{4 - a^{2}}}}^{\frac{2 \sqrt{2} - 2 + a}{\sqrt{4 - a^{2}}}} \frac{1}{1 + u^{2}} \times \frac{\sqrt{4 - a^{2}}}{2} d u$ $= \frac{2}{\sqrt{4 - a^{2}}} {[a r c t a n u]}_{\frac{a}{\sqrt{4 - a^{2}}}}^{\frac{2 \sqrt{2} - 2 + a}{\sqrt{4 - a^{2}}}} = \frac{2}{\sqrt{4 - a^{2}}} {a r c t a n (\frac{2 \sqrt{2} - 2 + a}{\sqrt{4 - a^{2}}})$ $- a r c t a n (\frac{a}{\sqrt{4 - a^{2}}})}$
Commented by mathmax by abdo last updated on 29/Feb/20

$2) w e h a v e f (a) = \int_{0}^{\frac{π}{4}} \frac{d x}{2 + a s i n x} \Rightarrow f^{^{'}} (a) = - \int_{0}^{\frac{π}{4}} \frac{s i n x d x}{{(2 + a s i n x)}^{2}} \Rightarrow$ $\int_{0}^{\frac{π}{4}} \frac{s i n x d x}{{(2 + a s i n x)}^{2}} = - f^{^{'}} (a) r e s t t h e c a l c u l u s o f f^{^{'}} (a) . . . b e c o n t i n u e d . . .$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum