find f(x) = ∫_(π/4) ^(π/3) ((cosxdx)/(2cos^2 x +sin^2 x +1))


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Question Number 42801 by maxmathsup by imad last updated on 02/Sep/18

$f i n d f (x) = \int_{\frac{π}{4}}^{\frac{π}{3}} \frac{c o s x d x}{2 {c o s}^{2} x + {s i n}^{2} x + 1}$
Commented by maxmathsup by imad last updated on 04/Sep/18
$let I = ∫_(π/4) ^(π/3) ((cosxdx)/(2(1−sin^2 x) +sin^2 x +1)) changement sinx =t give I = ∫_(1/(√2)) ^((√3)/2) (dt/(2(1−t^2 )+t^2 +1)) = ∫_(1/(√2)) ^((√3)/2) (dt/(3 −t^2 )) =∫_(1/(√2)) ^((√3)/2) (dt/(((√3)−t)((√3)+t)))dt = (1/(2(√3))) ∫_(1/(√2)) ^((√3)/2) {(1/((√3)−t)) +(1/((√3)+t))}dt =(1/(2(√3))) [ln∣(((√3)+t)/((√3)−t))∣]_(1/(√2)) ^((√3)/2) = (1/(2(√3))){ ln∣(((√3)+((√3)/2))/((√3)−((√3)/2)))∣ −ln∣(((√3)+(1/(√2)))/((√3)−(1/(√2))))∣} = (1/(2(√3))){ ln(((3(√3))/(√3)))−ln((((√6)+1)/((√6)−1)))} =(1/(2(√3))){ln(3)−ln(((1+(√6))/(−1+(√6))))} .$
$l e t I = \int_{\frac{π}{4}}^{\frac{π}{3}} \frac{c o s x d x}{2 (1 - {s i n}^{2} x) + {s i n}^{2} x + 1} c h a n g e m e n t s i n x = t g i v e$ $I = \int_{\frac{1}{\sqrt{2}}}^{\frac{\sqrt{3}}{2}} \frac{d t}{2 (1 - t^{2}) + t^{2} + 1} = \int_{\frac{1}{\sqrt{2}}}^{\frac{\sqrt{3}}{2}} \frac{d t}{3 - t^{2}} = \int_{\frac{1}{\sqrt{2}}}^{\frac{\sqrt{3}}{2}} \frac{d t}{(\sqrt{3} - t) (\sqrt{3} + t)} d t$ $= \frac{1}{2 \sqrt{3}} \int_{\frac{1}{\sqrt{2}}}^{\frac{\sqrt{3}}{2}} {\frac{1}{\sqrt{3} - t} + \frac{1}{\sqrt{3} + t}} d t = \frac{1}{2 \sqrt{3}} {[l n ∣ \frac{\sqrt{3} + t}{\sqrt{3} - t} ∣]}_{\frac{1}{\sqrt{2}}}^{\frac{\sqrt{3}}{2}} = \frac{1}{2 \sqrt{3}} {l n ∣ \frac{\sqrt{3} + \frac{\sqrt{3}}{2}}{\sqrt{3} - \frac{\sqrt{3}}{2}} ∣$ $- l n ∣ \frac{\sqrt{3} + \frac{1}{\sqrt{2}}}{\sqrt{3} - \frac{1}{\sqrt{2}}} ∣} = \frac{1}{2 \sqrt{3}} {l n (\frac{3 \sqrt{3}}{\sqrt{3}}) - l n (\frac{\sqrt{6} + 1}{\sqrt{6} - 1})} = \frac{1}{2 \sqrt{3}} {l n (3) - l n (\frac{1 + \sqrt{6}}{- 1 + \sqrt{6}})} .$
	Answered by tanmay.chaudhury50@gmail.com last updated on 03/Sep/18
	$∫_(1/(√2)) ^((√3)/2) (dt/(2(1−t^2 )+t^2 +1)) t=sinx dt=cosxdx ∫_(1/(√2)) ^((√3)/2) (dt/(3−t^2 ))=∣(1/(2(√3)))ln(((t+(√3))/(t−(√3))))∣_(1/(√2)) ^((√3)/2) (1/(2(√3))){ln∣(((((√3)/2)+(√3))/(((√3)/2)−(√3))))∣−ln∣((((1/(√2))+(√3))/((1/(√2))−(√3))))∣} =(1/(2(√3))){ln∣(((3/2)/(−(1/2))))∣−ln∣(((1+(√6))/(1−(√6))))∣} =(1/(2(√3))){ln∣−3∣−ln∣(((1+(√6))/(1−(√6))))} use formula ∫(dx/(a^2 −x^2 ))=(1/(2a))∫((a+x+a−x)/((a+x)(a−x)))dx (1/(2a))[∫(dx/(a−x))+∫(dx/(a+x))] (1/(2a))[∫(dx/(x+a))−∫(dx/(x−a))]=(1/(2a))ln(((x+a)/(x−a)))$
	$\int_{\frac{1}{\sqrt{2}}}^{\frac{\sqrt{3}}{2}} \frac{d t}{2 (1 - t^{2}) + t^{2} + 1} t = s i n x d t = c o s x d x$ $\int_{\frac{1}{\sqrt{2}}}^{\frac{\sqrt{3}}{2}} \frac{d t}{3 - t^{2}} =∣ \frac{1}{2 \sqrt{3}} l n (\frac{t + \sqrt{3}}{t - \sqrt{3}}) ∣_{\frac{1}{\sqrt{2}}}^{\frac{\sqrt{3}}{2}}$ $\frac{1}{2 \sqrt{3}} {l n ∣ (\frac{\frac{\sqrt{3}}{2} + \sqrt{3}}{\frac{\sqrt{3}}{2} - \sqrt{3}}) ∣ - l n ∣ (\frac{\frac{1}{\sqrt{2}} + \sqrt{3}}{\frac{1}{\sqrt{2}} - \sqrt{3}}) ∣}$ $= \frac{1}{2 \sqrt{3}} {l n ∣ (\frac{\frac{3}{2}}{- \frac{1}{2}}) ∣ - l n ∣ (\frac{1 + \sqrt{6}}{1 - \sqrt{6}}) ∣}$ $= \frac{1}{2 \sqrt{3}} {l n ∣ - 3 ∣ - l n ∣ (\frac{1 + \sqrt{6}}{1 - \sqrt{6}})}$ $u s e f o r m u l a \int \frac{d x}{a^{2} - x^{2}} = \frac{1}{2 a} \int \frac{a + x + a - x}{(a + x) (a - x)} d x$ $\frac{1}{2 a} [\int \frac{d x}{a - x} + \int \frac{d x}{a + x}]$ $\frac{1}{2 a} [\int \frac{d x}{x + a} - \int \frac{d x}{x - a}] = \frac{1}{2 a} l n (\frac{x + a}{x - a})$
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