calculate ∫_(π/4) ^(π/3) ((sinx)/(cosx +tanx))dx .


Question and Answers Forum

All Questions Topic List
Integration Questions
Previous in All Question Next in All Question
Previous in Integration Next in Integration
Question Number 42679 by maxmathsup by imad last updated on 31/Aug/18

$c a l c u l a t e \int_{\frac{π}{4}}^{\frac{π}{3}} \frac{s i n x}{c o s x + t a n x} d x .$
Commented by Meritguide1234 last updated on 03/Sep/18

Commented by maxmathsup by imad last updated on 03/Sep/18
$let A = ∫_(π/4) ^(π/3) ((sinx)/(cosx +tanx)) dx ⇒ A = ∫_(π/4) ^(π/3) ((sinx cosx)/(cos^2 x +sinx))dx = ∫_(π/4) ^(π/3) ((sinx cosxdx)/(1−sin^2 x +sinx)) =_(sinx =t) ∫_(1/(√2)) ^((√3)/2) ((t dt)/(1−t^2 +t)) = −∫_(1/(√2)) ^((√3)/2) (t/(t^2 −t −1)) dt =−(1/2) ∫_(1/(√2)) ^((√3)/2) ((2t−1−1)/(t^2 −t−1))dt =−(1/2)[ ln∣t^2 −t−1∣]_(1/(√2)) ^((√3)/2) +(1/2) ∫_(1/(√2)) ^((√3)/2) (dt/(t^2 −t −1)) =−(1/2){ ln∣(3/4)−((√3)/2)−1∣−ln∣(1/2)−(1/(√2))−1∣ +(1/2) ∫_(1/(√2)) ^((√3)/2) (dt/(t^2 −t −1)) but ∫_(1/(√2)) ^((√3)/2) (dt/(t^2 −t−1))dt = ∫_(1/(√2)) ^((√3)/2) (dt/((t−(1/2))^2 −(3/4))) =_(t−(1/2)=((√3)/2) u) ∫_((2/(√3))((1/(√2))−(1/2))) ^((2/(√3))((((√3)−1)/2))) (1/((3/4)(u^2 −1)))((√3)/2)du =(4/3) ((√3)/2) ∫_((1/(√3))((√2)−1)) ^(((√3)−1)/(√3)) (du/(u^2 −1)) =(1/(√3)) ∫_(((√2)−1)/(√3)) ^(((√3)−1)/(√3)) ( (1/(u−1)) −(1/(u+1)))du =(1/(√3))[ln∣((u−1)/(u+1))∣]_(((√2)−1)/(√3)) ^(((√3)−1)/(√3)) =(1/(√3)){ ln∣ (((((√3)−1)/(√3))−1)/((((√3)−1)/(√3))+1))∣ −ln∣(((((√2)−1)/(√3))−1)/((((√2)−1)/(√3)) +1))∣} =(1/(√3)){ ln∣ (1/(2(√3)−1))∣ −ln∣(((√2)−1−(√3))/((√2)−1 +(√3)))∣ ⇒ I =−(1/2)ln((1/4)+((√3)/2))−ln((1/2)+(1/(√2))) +(1/(√3)){ ln∣(1/(2(√3)−1))∣−ln∣(((√2)−1−(√3))/((√2)−1+(√3)))∣ .$
$l e t A = \int_{\frac{π}{4}}^{\frac{π}{3}} \frac{s i n x}{c o s x + t a n x} d x \Rightarrow A = \int_{\frac{π}{4}}^{\frac{π}{3}} \frac{s i n x c o s x}{{c o s}^{2} x + s i n x} d x$ $= \int_{\frac{π}{4}}^{\frac{π}{3}} \frac{s i n x c o s x d x}{1 - {s i n}^{2} x + s i n x} =_{s i n x = t} \int_{\frac{1}{\sqrt{2}}}^{\frac{\sqrt{3}}{2}} \frac{t d t}{1 - t^{2} + t}$ $= - \int_{\frac{1}{\sqrt{2}}}^{\frac{\sqrt{3}}{2}} \frac{t}{t^{2} - t - 1} d t = - \frac{1}{2} \int_{\frac{1}{\sqrt{2}}}^{\frac{\sqrt{3}}{2}} \frac{2 t - 1 - 1}{t^{2} - t - 1} d t$ $= - \frac{1}{2} {[l n ∣ t^{2} - t - 1 ∣]}_{\frac{1}{\sqrt{2}}}^{\frac{\sqrt{3}}{2}} + \frac{1}{2} \int_{\frac{1}{\sqrt{2}}}^{\frac{\sqrt{3}}{2}} \frac{d t}{t^{2} - t - 1}$ $= - \frac{1}{2} {l n ∣ \frac{3}{4} - \frac{\sqrt{3}}{2} - 1 ∣ - l n ∣ \frac{1}{2} - \frac{1}{\sqrt{2}} - 1 ∣ + \frac{1}{2} \int_{\frac{1}{\sqrt{2}}}^{\frac{\sqrt{3}}{2}} \frac{d t}{t^{2} - t - 1} b u t$ $\int_{\frac{1}{\sqrt{2}}}^{\frac{\sqrt{3}}{2}} \frac{d t}{t^{2} - t - 1} d t = \int_{\frac{1}{\sqrt{2}}}^{\frac{\sqrt{3}}{2}} \frac{d t}{{(t - \frac{1}{2})}^{2} - \frac{3}{4}} =_{t - \frac{1}{2} = \frac{\sqrt{3}}{2} u} \int_{\frac{2}{\sqrt{3}} (\frac{1}{\sqrt{2}} - \frac{1}{2})}^{\frac{2}{\sqrt{3}} (\frac{\sqrt{3} - 1}{2})} \frac{1}{\frac{3}{4} (u^{2} - 1)} \frac{\sqrt{3}}{2} d u$ $= \frac{4}{3} \frac{\sqrt{3}}{2} \int_{\frac{1}{\sqrt{3}} (\sqrt{2} - 1)}^{\frac{\sqrt{3} - 1}{\sqrt{3}}} \frac{d u}{u^{2} - 1} = \frac{1}{\sqrt{3}} \int_{\frac{\sqrt{2} - 1}{\sqrt{3}}}^{\frac{\sqrt{3} - 1}{\sqrt{3}}} (\frac{1}{u - 1} - \frac{1}{u + 1}) d u$ $= \frac{1}{\sqrt{3}} {[l n ∣ \frac{u - 1}{u + 1} ∣]}_{\frac{\sqrt{2} - 1}{\sqrt{3}}}^{\frac{\sqrt{3} - 1}{\sqrt{3}}} = \frac{1}{\sqrt{3}} {l n ∣ \frac{\frac{\sqrt{3} - 1}{\sqrt{3}} - 1}{\frac{\sqrt{3} - 1}{\sqrt{3}} + 1} ∣ - l n ∣ \frac{\frac{\sqrt{2} - 1}{\sqrt{3}} - 1}{\frac{\sqrt{2} - 1}{\sqrt{3}} + 1} ∣}$ $= \frac{1}{\sqrt{3}} {l n ∣ \frac{1}{2 \sqrt{3} - 1} ∣ - l n ∣ \frac{\sqrt{2} - 1 - \sqrt{3}}{\sqrt{2} - 1 + \sqrt{3}} ∣ \Rightarrow$ $I = - \frac{1}{2} l n (\frac{1}{4} + \frac{\sqrt{3}}{2}) - l n (\frac{1}{2} + \frac{1}{\sqrt{2}}) + \frac{1}{\sqrt{3}} {l n ∣ \frac{1}{2 \sqrt{3} - 1} ∣ - l n ∣ \frac{\sqrt{2} - 1 - \sqrt{3}}{\sqrt{2} - 1 + \sqrt{3}} ∣ .$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum