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Question Number 65740 by rajesh4661kumar@gmail.com last updated on 03/Aug/19

Commented by mathmax by abdo last updated on 03/Aug/19
$let A = ∫ (dx/((1/(cosx)) +sinx)) ⇒ A =∫ ((cosx)/(1+cosx sinx))dx changement tan((x/(2 ))) =t give A =∫ (((1−t^2 )/(1+t^2 ))/(1+((2t(1−t^2 ))/((1+t^2 )^2 )))) ((2dt)/(1+t^2 )) = ∫ ((1−t^2 )/((1+t^2 )^2 {1+((2t(1−t^2 ))/((1+t^2 )^2 ))}))dt =∫ ((1−t^2 )/((1+t^2 )^2 +2t(1−t^2 )))dt =∫ ((1−t^2 )/(t^4 +2t^2 +1 +2t−2t^3 ))dt =∫ ((1−t^2 )/(t^4 −2t^3 +2t +1))dt let decompose F(t) =((1−t^2 )/(t^4 −2t^3 +2t +1)) t^4 −2t^3 +2t +1 =0 the roots are z_1 =1,5291 +0,7429i =α+iβ z_2 =1,5291−0,7429 i (=z_1 ^− ) =α−iβ z_3 =−0,5291 +0,2571i =−α+iλ z_4 =−0,5291 −0,2571 i(=z_3 ^− ) =−α−iλ ⇒F(t)=((1−t^2 )/((t−z_1 )(t−z_1 ^− )(t−z_3 )(t−z_3 ^− ))) =((1−t^2 )/((t^2 −2Re(z_1 )t +∣z_1 ∣^2 )(t^2 −2Re(z_3 )t +∣z_3 ^2 ∣))) =((1−t^2 )/((t^2 −2αt +α^2 +β^2 )(t^2 +2αt +α^2 +λ^2 ))) =((at +b)/(t^2 −2αt +α^2 +β^2 )) +((ct +d)/(t^2 +2αt +α^2 +λ^2 )) ....be continued...$
$l e t A = \int \frac{d x}{\frac{1}{c o s x} + s i n x} \Rightarrow A = \int \frac{c o s x}{1 + c o s x s i n x} d x c h a n g e m e n t$ $t a n (\frac{x}{2}) = t g i v e A = \int \frac{\frac{1 - t^{2}}{1 + t^{2}}}{1 + \frac{2 t (1 - t^{2})}{{(1 + t^{2})}^{2}}} \frac{2 d t}{1 + t^{2}}$ $= \int \frac{1 - t^{2}}{{(1 + t^{2})}^{2} {1 + \frac{2 t (1 - t^{2})}{{(1 + t^{2})}^{2}}}} d t = \int \frac{1 - t^{2}}{{(1 + t^{2})}^{2} + 2 t (1 - t^{2})} d t$ $= \int \frac{1 - t^{2}}{t^{4} + 2 t^{2} + 1 + 2 t - 2 t^{3}} d t = \int \frac{1 - t^{2}}{t^{4} - 2 t^{3} + 2 t + 1} d t$ $l e t d e c o m p o s e F (t) = \frac{1 - t^{2}}{t^{4} - 2 t^{3} + 2 t + 1}$ $t^{4} - 2 t^{3} + 2 t + 1 = 0 t h e r o o t s a r e$ $z_{1} = 1, 5291 + 0, 7429 i = α + i β$ $z_{2} = 1, 5291 - 0, 7429 i (= {\bar{z}}_{1}) = α - i β$ $z_{3} = - 0, 5291 + 0, 2571 i = - α + i λ$ $z_{4} = - 0, 5291 - 0, 2571 i (= {\bar{z}}_{3}) = - α - i λ$ $\Rightarrow F (t) = \frac{1 - t^{2}}{(t - z_{1}) (t - {\bar{z}}_{1}) (t - z_{3}) (t - {\bar{z}}_{3})}$ $= \frac{1 - t^{2}}{(t^{2} - 2 R e (z_{1}) t + ∣ z_{1} ∣^{2}) (t^{2} - 2 R e (z_{3}) t + ∣ z_{3}^{2} ∣)}$ $= \frac{1 - t^{2}}{(t^{2} - 2 α t + α^{2} + β^{2}) (t^{2} + 2 α t + α^{2} + λ^{2})}$ $= \frac{a t + b}{t^{2} - 2 α t + α^{2} + β^{2}} + \frac{c t + d}{t^{2} + 2 α t + α^{2} + λ^{2}} . . . . b e c o n t i n u e d . . .$
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