calculate ∫_0 ^(π/4) (dx/((cosx +3sinx)^2 ))


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Question Number 84578 by msup trace by abdo last updated on 14/Mar/20

$c a l c u l a t e \int_{0}^{\frac{π}{4}} \frac{d x}{{(c o s x + 3 s i n x)}^{2}}$
Commented by jagoll last updated on 14/Mar/20

$\int \underset{0}{\overset{\frac{π}{4}}{}} \frac{dx}{(\sqrt{10} \cos {(x - \tan^{- 1} (3))}^{2}}$ $= \frac{1}{10} \int \underset{0}{\overset{\frac{π}{4}}{}} \sec^{2} (x - \tan^{- 1} (3)) dx$ $= \frac{1}{10} \tan (x - \tan^{- 1} (3))]_{0}^{\frac{π}{4}}$ $= \frac{1}{10} [\tan (\frac{π}{4} - \tan^{- 1} (3)) - \tan (- \tan^{- 1} (3))]$
Commented by jagoll last updated on 14/Mar/20

$= \frac{1}{10} [\frac{1 - 3}{1 + 3} + 3] = \frac{1}{10} [\frac{5}{2}] = \frac{1}{4}$
Commented by abdomathmax last updated on 14/Mar/20
$let f(a)=∫_0 ^(π/4) (dx/(a +cosx +3sinx)) ⇒ f^′ (a)=−∫_0 ^(π/4) (dx/((a +cosx +3sinx)^2 )) ⇒ ∫_0 ^(π/4) (dx/((cosx +3sinx)^2 ))=−f^′ (0) let explicit f(a) f(a) =_(tan((x/2))=t) ∫_0 ^((√2)−1) (1/(a+((1−t^2 )/(1+t^2 ))+3((2t)/(1+t^2 ))))((2dt)/(1+t^2 )) = ∫_0 ^((√2)−1) ((2dt)/(a(1+t^2 )+1−t^2 +6t)) =∫_0 ^((√2)−1) (dt/((a−1)t^2 +6t +a+1)) Δ^′ =9−(a−1)(a+1) =9−(a^2 −1)=10−a^2 t_1 =((−3+(√(10−a^2 )))/(a−1)) and t_2 =((−3−(√(10−a^2 )))/(a−1)) (a≠1) f(a) =∫_0 ^((√2)−1) ((2dt)/((a−1)(t−t_1 )(t−t_2 ))) =(2/(a−1)) (1/(t_1 −t_2 ))∫_0 ^((√2)−1) {(1/(t−t_1 ))−(1/(t−t_2 ))}dt =(2/(a−1))×(1/((2(√(10−a^2 )))/(a−1)))[ln∣((t−t_1 )/(t−t_2 ))∣]_0 ^((√2)−1) =(1/(√(10−a^2 ))){ln∣(((√2)−1−t_1 )/((√2)−1−t_2 ))∣−ln∣(t_1 /t_2 )∣} =(1/(√(10−a^2 ))){ln∣(((√2)−1−((−3+(√(10−a^2 )))/(a−1)))/((√2)−1+((3+(√(10−a^2 )))/(a−1))))∣ −ln∣((−3+(√(10−a^2 )))/(−3−(√(10−a^2 ))))∣$
$l e t f (a) = \int_{0}^{\frac{π}{4}} \frac{d x}{a + c o s x + 3 s i n x} \Rightarrow$ $f^{^{'}} (a) = - \int_{0}^{\frac{π}{4}} \frac{d x}{{(a + c o s x + 3 s i n x)}^{2}} \Rightarrow$ $\int_{0}^{\frac{π}{4}} \frac{d x}{{(c o s x + 3 s i n x)}^{2}} = - f^{^{'}} (0) l e t e x p l i c i t f (a)$ $f (a) =_{t a n (\frac{x}{2}) = t} \int_{0}^{\sqrt{2} - 1} \frac{1}{a + \frac{1 - t^{2}}{1 + t^{2}} + 3 \frac{2 t}{1 + t^{2}}} \frac{2 d t}{1 + t^{2}}$ $= \int_{0}^{\sqrt{2} - 1} \frac{2 d t}{a (1 + t^{2}) + 1 - t^{2} + 6 t}$ $= \int_{0}^{\sqrt{2} - 1} \frac{d t}{(a - 1) t^{2} + 6 t + a + 1}$ $Δ^{^{'}} = 9 - (a - 1) (a + 1) = 9 - (a^{2} - 1) = 10 - a^{2}$ $t_{1} = \frac{- 3 + \sqrt{10 - a^{2}}}{a - 1} a n d t_{2} = \frac{- 3 - \sqrt{10 - a^{2}}}{a - 1} (a \neq 1)$ $f (a) = \int_{0}^{\sqrt{2} - 1} \frac{2 d t}{(a - 1) (t - t_{1}) (t - t_{2})}$ $= \frac{2}{a - 1} \frac{1}{t_{1} - t_{2}} \int_{0}^{\sqrt{2} - 1} {\frac{1}{t - t_{1}} - \frac{1}{t - t_{2}}} d t$ $= \frac{2}{a - 1} \times \frac{1}{\frac{2 \sqrt{10 - a^{2}}}{a - 1}} {[l n ∣ \frac{t - t_{1}}{t - t_{2}} ∣]}_{0}^{\sqrt{2} - 1}$ $= \frac{1}{\sqrt{10 - a^{2}}} {l n ∣ \frac{\sqrt{2} - 1 - t_{1}}{\sqrt{2} - 1 - t_{2}} ∣ - l n ∣ \frac{t_{1}}{t_{2}} ∣}$ $= \frac{1}{\sqrt{10 - a^{2}}} {l n ∣ \frac{\sqrt{2} - 1 - \frac{- 3 + \sqrt{10 - a^{2}}}{a - 1}}{\sqrt{2} - 1 + \frac{3 + \sqrt{10 - a^{2}}}{a - 1}} ∣$ $- l n ∣ \frac{- 3 + \sqrt{10 - a^{2}}}{- 3 - \sqrt{10 - a^{2}}} ∣$
Commented by abdomathmax last updated on 14/Mar/20
$f(a) =(1/(√(10−a^2 ))){ln∣((((√2)−1)(a−1)+3−(√(10−a^2 )))/(((√2)−1)(a−1)+3+(√(10−a^2 ))))∣ −ln∣((3−(√(10−a^2 )))/(3+(√(10−a^2 ))))∣ rest calculus of f^′ (a) and f^′ (0) ...$
$f (a) = \frac{1}{\sqrt{10 - a^{2}}} {l n ∣ \frac{(\sqrt{2} - 1) (a - 1) + 3 - \sqrt{10 - a^{2}}}{(\sqrt{2} - 1) (a - 1) + 3 + \sqrt{10 - a^{2}}} ∣$ $- l n ∣ \frac{3 - \sqrt{10 - a^{2}}}{3 + \sqrt{10 - a^{2}}} ∣ r e s t c a l c u l u s o f f^{^{'}} (a) a n d f^{^{'}} (0)$ $. . .$
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