∫(1/(2cosx−5sinx−3))dx


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Question Number 71239 by 20190927 last updated on 13/Oct/19

$\int \frac{1}{2 cosx - 5 sinx - 3} dx$
Commented by mathmax by abdo last updated on 13/Oct/19
$let I =∫ (dx/(2cosx−5sinx −3)) changement tan((x/2))=t give I =∫ (1/(2((1−t^2 )/(1+t^2 ))−5((2t)/(1+t^2 ))−3))((2dt)/(1+t^2 )) =∫((2dt)/(2−2t^2 −10t−3−3t^2 )) =∫ ((2dt)/(−5t^2 −10t−1)) =∫((−2dt)/(5t^2 +10t +1)) 5t^2 +10t +1=0→Δ^′ =5^2 −5 =25−5=20 ⇒ t_1 =((−5+2(√5))/5) =−1+(2/(√5)) and t_2 =−1−(2/(√5)) ⇒ ∫ (dt/(5t^2 +10t+1)) =∫ (dt/(5(t−t_1 )(t−t_2 ))) =(1/(5(t_1 −t_2 )))∫ ((1/(t−t_1 ))−(1/(t−t_2 )))dt =(1/(5(4/(√5)))) ∫ ((1/(t−t_1 ))−(1/(t−t_2 )))dt =(1/(4(√5))){ln∣t−t_1 ∣−ln∣t−t_2 ∣} +c ⇒I=−(1/(2(√5))){ln∣t−t_1 ∣−ln∣t−t_2 ∣} +c =−(1/(2(√5)))ln∣tan((x/2))+1−(2/(√5))∣+(1/(2(√5)))ln∣tan((x/2))+1+(2/(√5))∣ +C$
$l e t I = \int \frac{d x}{2 c o s x - 5 s i n x - 3} c h a n g e m e n t t a n (\frac{x}{2}) = t g i v e$ $I = \int \frac{1}{2 \frac{1 - t^{2}}{1 + t^{2}} - 5 \frac{2 t}{1 + t^{2}} - 3} \frac{2 d t}{1 + t^{2}} = \int \frac{2 d t}{2 - 2 t^{2} - 10 t - 3 - 3 t^{2}}$ $= \int \frac{2 d t}{- 5 t^{2} - 10 t - 1} = \int \frac{- 2 d t}{5 t^{2} + 10 t + 1}$ $5 t^{2} + 10 t + 1 = 0 \to Δ^{^{'}} = 5^{2} - 5 = 25 - 5 = 20 \Rightarrow$ $t_{1} = \frac{- 5 + 2 \sqrt{5}}{5} = - 1 + \frac{2}{\sqrt{5}} a n d t_{2} = - 1 - \frac{2}{\sqrt{5}} \Rightarrow$ $\int \frac{d t}{5 t^{2} + 10 t + 1} = \int \frac{d t}{5 (t - t_{1}) (t - t_{2})} = \frac{1}{5 (t_{1} - t_{2})} \int (\frac{1}{t - t_{1}} - \frac{1}{t - t_{2}}) d t$ $= \frac{1}{5 \frac{4}{\sqrt{5}}} \int (\frac{1}{t - t_{1}} - \frac{1}{t - t_{2}}) d t = \frac{1}{4 \sqrt{5}} {l n ∣ t - t_{1} ∣ - l n ∣ t - t_{2} ∣} + c$ $\Rightarrow I = - \frac{1}{2 \sqrt{5}} {l n ∣ t - t_{1} ∣ - l n ∣ t - t_{2} ∣} + c$ $= - \frac{1}{2 \sqrt{5}} l n ∣ t a n (\frac{x}{2}) + 1 - \frac{2}{\sqrt{5}} ∣ + \frac{1}{2 \sqrt{5}} l n ∣ t a n (\frac{x}{2}) + 1 + \frac{2}{\sqrt{5}} ∣ + C$
Commented by 20190927 last updated on 13/Oct/19

$thank you very much$
Commented by turbo msup by abdo last updated on 13/Oct/19

$y o u a r e w e l c o m e$
	Answered by MJS last updated on 13/Oct/19

	$\int \frac{d x}{2 \cos x - 5 \sin x - 3} =$ $[t = \tan \frac{x}{2} \to d x = \frac{2 d t}{t^{2} + 1}]$ $= - 2 \int \frac{d t}{5 t^{2} + 10 t + 1} =$ $= \frac{\sqrt{5}}{2} \int \frac{d t}{5 t + 5 + 2 \sqrt{5}} - \frac{\sqrt{5}}{2} \int \frac{d t}{5 t + 5 - 2 \sqrt{5}} =$ $= \frac{\sqrt{5}}{10} \ln \frac{5 t + 5 + 2 \sqrt{5}}{5 t + 5 - 2 \sqrt{5}} = . . .$
	Answered by peter frank last updated on 13/Oct/19

	$t = \tan \frac{x}{2}$ $d x = \frac{2}{1 + t^{2}} d t$ $2 (\frac{1 - t^{2}}{1 + t^{2}}) - 5 (\frac{2 t}{1 + t^{2}}) - 3$ $\frac{2 - 2 t^{2} - 10 t - 3 (1 + t^{2})}{1 + t^{2}}$ $\frac{2 - 2 t^{2} - 10 t - 3 - 3 t^{2}}{1 + t^{2}}$ $\frac{- 5 t^{2} - 10 t - 5}{1 + t^{2}}$ $\int \frac{\frac{2}{1 + t^{2}}}{\frac{- 5 t^{2} - 10 t - 5}{1 + t^{2}}} d t$ $\int \frac{2 d t}{t^{2} + 2 t + 1}$ $\int \frac{2 d t}{{(t + 1)}^{2} - {(\frac{2}{\sqrt{5}})}^{2}}$ $. . . . . . . .$
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