Math Question and Answers


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 41634 by Tawa1 last updated on 10/Aug/18

Commented by maxmathsup by imad last updated on 10/Aug/18
$I = ∫ (dx/(sinx +(1/(cosx)))) = ∫ ((cosx)/(sinx .cosx +1))dx changement tan((x/2))=t give I = ∫ (((1−t^2 )/(1+t^2 ))/(((2t(1−t^2 ))/((1+t^2 )^2 )) +1)) ((2dt)/(1+t^2 )) = ∫ (((1−t^2 )/(1+t^2 ))/(((2t(1−t^2 ))/(1 +t^2 )) +1+t^2 )) 2dt = ∫ ((2(1−t^2 ))/(2t(1−t^2 ) +(1+t^2 )^2 ))dt = ∫ ((2−2t^2 )/(2t−2t^3 +t^4 +2t^2 +1)) dt = ∫ ((−2t^2 +2)/(t^4 −2t^3 +2t^2 +2t +1)) dt the roots of p(t)=t^4 −2t^3 +2t^2 +2t +1 are t_1 ∼ 1,366 +1,366 i =α(1+i) t_2 ∼ 1,366−1,366 i =α(1−i) with α =1,366 t_3 ∼−0,366 +0,366 i =β(1−i) and t_4 =β(1+i) with β =−0,366 t_4 =−0,366 −0,366i all roots are complex so F(t) = ((−2t^2 +2)/((t −α(1+i))(t−α(1−i))(t −β(1−i)(t−β(1+i))) = ((−2t^2 +2)/({(t−α)^2 +α^2 }{ (t−β)^2 +β^2 })) =((at +b)/((t−α)^2 +α^2 )) +((ct +d)/((t−β)^2 +β^2 )) ⇒ ∫ F(t)dt = ∫ ((at+b)/((t−α)^2 +α^2 )) dt + ∫ ((ct +d)/((t−β)^2 +β^2 )) dt +c... be continued...$
$I = \int \frac{d x}{s i n x + \frac{1}{c o s x}} = \int \frac{c o s x}{s i n x . c o s x + 1} d x c h a n g e m e n t t a n (\frac{x}{2}) = t g i v e$ $I = \int \frac{\frac{1 - t^{2}}{1 + t^{2}}}{\frac{2 t (1 - t^{2})}{{(1 + t^{2})}^{2}} + 1} \frac{2 d t}{1 + t^{2}} = \int \frac{\frac{1 - t^{2}}{1 + t^{2}}}{\frac{2 t (1 - t^{2})}{1 + t^{2}} + 1 + t^{2}} 2 d t$ $= \int \frac{2 (1 - t^{2})}{2 t (1 - t^{2}) + {(1 + t^{2})}^{2}} d t = \int \frac{2 - 2 t^{2}}{2 t - 2 t^{3} + t^{4} + 2 t^{2} + 1} d t$ $= \int \frac{- 2 t^{2} + 2}{t^{4} - 2 t^{3} + 2 t^{2} + 2 t + 1} d t t h e r o o t s o f p (t) = t^{4} - 2 t^{3} + 2 t^{2} + 2 t + 1 a r e$ $t_{1} \sim 1, 366 + 1, 366 i = α (1 + i)$ $t_{2} \sim 1, 366 - 1, 366 i = α (1 - i) w i t h α = 1, 366$ $t_{3} \sim - 0, 366 + 0, 366 i = β (1 - i) a n d t_{4} = β (1 + i) w i t h β = - 0, 366$ $t_{4} = - 0, 366 - 0, 366 i a l l r o o t s a r e c o m p l e x s o$ $F (t) = \frac{- 2 t^{2} + 2}{(t - α (1 + i)) (t - α (1 - i)) (t - β (1 - i) (t - β (1 + i)}$ $= \frac{- 2 t^{2} + 2}{{{(t - α)}^{2} + α^{2}} {{(t - β)}^{2} + β^{2}}} = \frac{a t + b}{{(t - α)}^{2} + α^{2}} + \frac{c t + d}{{(t - β)}^{2} + β^{2}} \Rightarrow$ $\int F (t) d t = \int \frac{a t + b}{{(t - α)}^{2} + α^{2}} d t + \int \frac{c t + d}{{(t - β)}^{2} + β^{2}} d t + c . . .$ $b e c o n t i n u e d . . .$
	Answered by tanmay.chaudhury50@gmail.com last updated on 10/Aug/18
	$∫((2cosx)/(2sinxcosx+2))dx ∫((cosx+sinx+cosx−sinx)/(1+1+2sinxcosx))dx ∫((cosx+sinx+cosx−sinx)/(1+cos^2 x+sin^2 x+2sinxcosx))dx ∫((cosx+sinx+cosx−sinx)/(1+(sinx+cosx)^2 ))dx ∫((cosx−sinx)/(1+(sinx+cosx)^2 ))dx+∫((−(cosx+sinx))/(−1−(sin^2 x+cos^2 x+2sinxcosx))dx ∫((d(sinx+cosx))/(1+(sinx+cosx)^2 ))−∫(((cosx+sinx)/(−1−1−2sinxcosx))dx ∫(dt/(1+t^2 ))−∫((cosx+sinx)/(−3+1−2sinxcosx))dx tan^(−1) (t)−∫((d(sinx−cosx))/(−3+(sinx−cosx)^2 )) tan^(−1) (sinx+cosx)−∫(dt_2 /(t_2 ^2 −((√3) )^2 )) ∫(dt_2 /((t_2 +(√3) )(t_2 −(√(3))))) =(1/(2(√3)))∫(((t_2 +(√3) )−(t_2 −(√3) ))/((t_2 +(√3) )(t−(√3) )))dt_2 (1/(2(√3))){∫(dt_2 /(t_2 −(√3) ))−∫(dt_2 /(t_2 +(√3) ))} (1/(2(√3) ))ln(((t_2 −(√3) )/(t_(2 ) +(√3))))+c ans is tan^(−1) (sinx+cosx)−(1/(2(√3) ))ln(((sinx−cosx−(√3))/(sinx−cosx+(√3))))+c$
	$\int \frac{2 c o s x}{2 s i n x c o s x + 2} d x$ $\int \frac{c o s x + s i n x + c o s x - s i n x}{1 + 1 + 2 s i n x c o s x} d x$ $\int \frac{c o s x + s i n x + c o s x - s i n x}{1 + {c o s}^{2} x + {s i n}^{2} x + 2 s i n x c o s x} d x$ $\int \frac{c o s x + s i n x + c o s x - s i n x}{1 + {(s i n x + c o s x)}^{2}} d x$ $\int \frac{c o s x - s i n x}{1 + {(s i n x + c o s x)}^{2}} d x + \int \frac{- (c o s x + s i n x)}{- 1 - ({s i n}^{2} x + {c o s}^{2} x + 2 s i n x c o s x} d x$ $\int \frac{d (s i n x + c o s x)}{1 + {(s i n x + c o s x)}^{2}} - \int \frac{(c o s x + s i n x}{- 1 - 1 - 2 s i n x c o s x} d x$ $\int \frac{d t}{1 + t^{2}} - \int \frac{c o s x + s i n x}{- 3 + 1 - 2 s i n x c o s x} d x$ ${t a n}^{- 1} (t) - \int \frac{d (s i n x - c o s x)}{- 3 + {(s i n x - c o s x)}^{2}}$ ${t a n}^{- 1} (s i n x + c o s x) - \int \frac{{d t}_{2}}{t_{2}^{2} - {(\sqrt{3})}^{2}}$ $\int \frac{{d t}_{2}}{(t_{2} + \sqrt{3}) (t_{2} - \sqrt{3)}}$ $= \frac{1}{2 \sqrt{3}} \int \frac{(t_{2} + \sqrt{3}) - (t_{2} - \sqrt{3})}{(t_{2} + \sqrt{3}) (t - \sqrt{3})} {d t}_{2}$ $\frac{1}{2 \sqrt{3}} {\int \frac{{d t}_{2}}{t_{2} - \sqrt{3}} - \int \frac{{d t}_{2}}{t_{2} + \sqrt{3}}}$ $\frac{1}{2 \sqrt{3}} l n (\frac{t_{2} - \sqrt{3}}{t_{2} + \sqrt{3}}) + c$ $a n s i s$ ${t a n}^{- 1} (s i n x + c o s x) - \frac{1}{2 \sqrt{3}} l n (\frac{s i n x - c o s x - \sqrt{3}}{s i n x - c o s x + \sqrt{3}}) + c$
	Commented by Tawa1 last updated on 10/Aug/18

	$God bless you sir$
	Answered by Meritguide1234 last updated on 12/Aug/18

	$= \int \frac{c o s x}{s i n x c o s x + 1} d x$ $= \int \frac{2 c o s x}{2 + s i n 2 x} d x$ $= \int \frac{(c o s x + s i n x) + (c o s x - s i n x)}{2 + s i n 2 x} d x$ $= \int \frac{d (s i n x - c o s x)}{3 - {(s i n x - c o s x)}^{2}} + \int \frac{d (c o s x + s i n x)}{1 + {(c o s x + s i n x)}^{2}}$
	Commented by Tawa1 last updated on 12/Aug/18

	$God bless you sir$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum