(1)∫(1/(cos (√x))) dx (2) ∫ (1/(2+cot x)) dx (3) ∫ (1/(ln(cos x))) dx


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Question Number 102690 by bramlex last updated on 10/Jul/20

$(1) \int \frac{1}{\cos \sqrt{x}} dx$ $(2) \int \frac{1}{2 + \cot x} dx$ $(3) \int \frac{1}{\ln (\cos x)} dx$
	Answered by Dwaipayan Shikari last updated on 10/Jul/20
	$2)∫((tanx)/(2tanx+1))=(1/2)∫((2tanx+1)/(2tanx+1))−(1/2)∫(1/(2tanx+1))=(x/2)−(1/2)∫(2/((2t+1)(t^2 +1)))dt =(x/2)−∫((At+C)/(t^2 +1))+(B/(2t+1)) {(1/((2t+1)(t^2 +1)))=((At+C)/(t^2 +1))+(B/(2t+1)) and take (t=tanx) =(x/2)−I_a {2At^2 +At+2Ct+C+Bt^2 +B=1 =(x/2)+∫(((2/5)t−(1/5))/(t^2 +1))−((4/5)/(2t+1)) { 2A+B=0 A+2C=0 B+C=1 =(x/2)+(1/5)∫((2t−1)/(t^2 +1))−(4/(10))∫(1/(t+(1/2)))dt {A=−(2/5) B=(4/5) C=(1/5) =(x/2)+(1/5)∫((2t)/(t^2 +1))−(1/5)tan^(−1) tanx−(4/(10))log(t+(1/2)) =(x/2)+(1/5)log(t^2 +1)−(1/5)x−(4/(10))log(tanx+(1/2)) =(x/2)+(2/5)log(secx)−(1/5)x−(4/(10))log(tanx+(1/2))+Constant =(1/5)(2x−log(cosx)−2log(tanx+(1/2)))+Constant$
	$2) \int \frac{t a n x}{2 t a n x + 1} = \frac{1}{2} \int \frac{2 t a n x + 1}{2 t a n x + 1} - \frac{1}{2} \int \frac{1}{2 t a n x + 1} = \frac{x}{2} - \frac{1}{2} \int \frac{2}{(2 t + 1) (t^{2} + 1)} d t$ $= \frac{x}{2} - \int \frac{A t + C}{t^{2} + 1} + \frac{B}{2 t + 1} {\frac{1}{(2 t + 1) (t^{2} + 1)} = \frac{A t + C}{t^{2} + 1} + \frac{B}{2 t + 1} a n d t a k e$ $(t = t a n x)$ $= \frac{x}{2} - I_{a} {2 {A t}^{2} + A t + 2 C t + C + {B t}^{2} + B = 1$ $= \frac{x}{2} + \int \frac{\frac{2}{5} t - \frac{1}{5}}{t^{2} + 1} - \frac{\frac{4}{5}}{2 t + 1} {2 A + B = 0 A + 2 C = 0 B + C = 1$ $= \frac{x}{2} + \frac{1}{5} \int \frac{2 t - 1}{t^{2} + 1} - \frac{4}{10} \int \frac{1}{t + \frac{1}{2}} d t {A = - \frac{2}{5} B = \frac{4}{5} C = \frac{1}{5}$ $= \frac{x}{2} + \frac{1}{5} \int \frac{2 t}{t^{2} + 1} - \frac{1}{5} {t a n}^{- 1} t a n x - \frac{4}{10} l o g (t + \frac{1}{2})$ $= \frac{x}{2} + \frac{1}{5} l o g (t^{2} + 1) - \frac{1}{5} x - \frac{4}{10} l o g (t a n x + \frac{1}{2})$ $= \frac{x}{2} + \frac{2}{5} l o g (s e c x) - \frac{1}{5} x - \frac{4}{10} l o g (t a n x + \frac{1}{2}) + C o n s t a n t$ $= \frac{1}{5} (2 x - l o g (c o s x) - 2 l o g (t a n x + \frac{1}{2})) + C o n s t a n t$
	Answered by mathmax by abdo last updated on 10/Jul/20

	$2) I = \int \frac{dx}{2 + \frac{cosx}{sinx}} \Rightarrow I = \int \frac{sinx}{2 sinx + cosx} dx changement \tan (\frac{x}{2}) = t give$ $I = \int \frac{\frac{2 t}{1 + t^{2}}}{2 \times \frac{2 t}{1 + t^{2}} + \frac{1 - t^{2}}{1 + t^{2}}} dt = \int \frac{2 t}{4 t + 1 - t^{2}} dt = - 2 \int \frac{tdt}{t^{2} - 4 t - 1}$ $t^{2} - 4 t - 1 = 0 \to Δ^{^{'}} = 4 + 1 = 5 \Rightarrow t_{1} = 2 + \sqrt{5} and t_{2} = 2 - \sqrt{5}$ $\Rightarrow F (t) = \frac{t}{t^{2} - 4 t - 1} = \frac{t}{(t - t_{1}) (t - t_{2})} = \frac{a}{t - t_{1}} + \frac{b}{t - t_{2}}$ $a = \frac{t_{1}}{t_{1} - t_{2}} = \frac{2 + \sqrt{5}}{2 \sqrt{5}}, b = \frac{t_{2}}{t_{2} - t_{1}} = \frac{2 - \sqrt{5}}{- 2 \sqrt{5}} = \frac{- 2 + \sqrt{5}}{2 \sqrt{5}} \Rightarrow$ $I = - 2 \int \frac{adt}{t - t_{1}} - 2 \int \frac{bdt}{t - t_{2}}$ $= - 2 aln ∣ t - t_{1} ∣ - 2 b \ln ∣ t - t_{2} ∣ + c$ $= - 2 \times \frac{2 + \sqrt{5}}{2 % 5} \ln ∣ \tan (\frac{x}{2}) - 2 - \sqrt{5} ∣ - 2 \times \frac{- 2 + \sqrt{5}}{2 \sqrt{5}} \ln ∣ \tan (\frac{x}{2}) - 2 + \sqrt{5} ∣ + C$ $= (\frac{- 2 - \sqrt{5}}{\sqrt{5}}) \ln ∣ \tan (\frac{x}{2}) - 2 - \sqrt{5} ∣ + \frac{2 - \sqrt{5}}{\sqrt{5}} \ln ∣ \tan (\frac{x}{2}) - 2 + \sqrt{5} ∣ + C$
	Answered by bemath last updated on 11/Jul/20

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