∫ ((sin x+cos x)/(3sin x+4cos x+1)) dx


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Question Number 126753 by bemath last updated on 24/Dec/20

$\int \frac{\sin x + \cos x}{3 \sin x + 4 \cos x + 1} d x$
	Answered by Ar Brandon last updated on 24/Dec/20
	$sinx+cosx=λ(3sinx+4cosx+1)+μ{(d/dx)(3sinx+4cosx+1)}+γ =λ(3sinx+4cosx+1)+μ(3cosx−4sinx)+γ { ((3λ−4μ=1)),((4λ+3μ=1)),((λ+γ=0)) :}⇒ { ((25μ=−1)),((25λ=7)),((γ=−λ)) :} I=∫((sinx+cosx)/(3sinx+4cosx+1))dx =(7/(25))∫((3sinx+4cosx+1)/(3sinx+4cosx+1))dx−(1/(25))∫((3cosx−4sinx)/(3sinx+4cosx+1))dx −(7/(25))∫(dx/(3sinx+4cosx+1)) =(7/(25))x−(1/(25))ln∣3sinx+4cosx+1∣−(7/(25))J tan(x/2)=t ⇒ sec^2 (x/2)dx=2tdt ⇒ dx=((2t)/(t^2 +1))dt J=∫(1/(((3t)/(t^2 +1))−4((t^2 −1)/(t^2 +1))+1))∙((2t)/(t^2 +1))dt=∫((2t)/(3t−4(t^2 −1)+(t^2 +1)))dt =∫((2t)/(−3t^2 +3t+5))dt=−∫((2t)/(3t^2 −3t−5))dt 2t=μ{(d/dt)(3t^2 −3t−5)}+γ=μ(6t−3)+γ { ((6μ=2)),((γ−3μ=0)) :} ⇒ { ((μ=(1/3))),((γ=1)) :} J=−∫{(1/3)∙((6t−3)/(3t^2 −3t−5))+(1/(3t^2 −3t−5))}dt =−{(1/3)ln∣3t^2 −3t−5∣−((2(√3))/( 3(√(23))))Arctanh((2t−1)(√(3/(23))))}+C =−((ln∣3t^2 −3t−5∣)/3)+(1/( (√(69))))ln∣((2t−1+3/(√(69)))/(2t−1−3/(√(69))))∣+C I=(7/(25))x−((ln∣3sinx+4cosx+1∣)/(25))+(7/(75))ln∣3tan^2 ((x/2))−3tan((x/2))−5∣ −(7/( 25(√(69))))ln∣((2tan(x/2)−1+3/(√(69)))/(2tan(x/2)−1−3/(√(69))))∣+C$
	$sinx + cosx = λ (3 sinx + 4 cosx + 1) + μ {\frac{d}{dx} (3 sinx + 4 cosx + 1)} + γ$ $= λ (3 sinx + 4 cosx + 1) + μ (3 cosx - 4 sinx) + γ$ ${\begin{cases} 3 λ - 4 μ = 1 \\ 4 λ + 3 μ = 1 \\ λ + γ = 0 \end{cases} \Rightarrow {\begin{cases} 25 μ = - 1 \\ 25 λ = 7 \\ γ = - λ \end{cases}$ $I = \int \frac{sinx + cosx}{3 sinx + 4 cosx + 1} dx$ $= \frac{7}{25} \int \frac{3 sinx + 4 cosx + 1}{3 sinx + 4 cosx + 1} dx - \frac{1}{25} \int \frac{3 cosx - 4 sinx}{3 sinx + 4 cosx + 1} dx$ $- \frac{7}{25} \int \frac{dx}{3 sinx + 4 cosx + 1}$ $= \frac{7}{25} x - \frac{1}{25} \ln ∣ 3 sinx + 4 cosx + 1 ∣ - \frac{7}{25} J$ $\tan \frac{x}{2} = t \Rightarrow \sec^{2} \frac{x}{2} dx = 2 tdt \Rightarrow dx = \frac{2 t}{t^{2} + 1} dt$ $J = \int \frac{1}{\frac{3 t}{t^{2} + 1} - 4 \frac{t^{2} - 1}{t^{2} + 1} + 1} \cdot \frac{2 t}{t^{2} + 1} dt = \int \frac{2 t}{3 t - 4 (t^{2} - 1) + (t^{2} + 1)} dt$ $= \int \frac{2 t}{- {3 t}^{2} + 3 t + 5} dt = - \int \frac{2 t}{{3 t}^{2} - 3 t - 5} dt$ $2 t = μ {\frac{d}{dt} ({3 t}^{2} - 3 t - 5)} + γ = μ (6 t - 3) + γ$ ${\begin{cases} 6 μ = 2 \\ γ - 3 μ = 0 \end{cases} \Rightarrow {\begin{cases} μ = \frac{1}{3} \\ γ = 1 \end{cases}$ $J = - \int {\frac{1}{3} \cdot \frac{6 t - 3}{{3 t}^{2} - 3 t - 5} + \frac{1}{{3 t}^{2} - 3 t - 5}} dt$ $= - {\frac{1}{3} \ln ∣ {3 t}^{2} - 3 t - 5 ∣ - \frac{2 \sqrt{3}}{3 \sqrt{23}} Arctanh ((2 t - 1) \sqrt{\frac{3}{23}})} + C$ $= - \frac{\ln ∣ {3 t}^{2} - 3 t - 5 ∣}{3} + \frac{1}{\sqrt{69}} \ln ∣ \frac{2 t - 1 + 3 / \sqrt{69}}{2 t - 1 - 3 / \sqrt{69}} ∣ + C$ $I = \frac{7}{25} x - \frac{\ln ∣ 3 sinx + 4 cosx + 1 ∣}{25} + \frac{7}{75} \ln ∣ {3 \tan}^{2} (\frac{x}{2}) - 3 \tan (\frac{x}{2}) - 5 ∣$ $- \frac{7}{25 \sqrt{69}} \ln ∣ \frac{2 \tan (x / 2) - 1 + 3 / \sqrt{69}}{2 \tan (x / 2) - 1 - 3 / \sqrt{69}} ∣ + C$
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