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Question Number 103863 by mohammad17 last updated on 17/Jul/20

	Answered by Dwaipayan Shikari last updated on 17/Jul/20

	$2)$ $\int \sqrt{\frac{1 + x}{1 - x}} d x = \int \frac{1 + x}{\sqrt{1 - x^{2}}} d x = \int \frac{1}{\sqrt{1 - x^{2}}} d x - \frac{1}{2} . \frac{- 2 x}{\sqrt{1 - x^{2}}} d x$ $= {s i n}^{- 1} x - \sqrt{1 - x^{2}} + C$
	Answered by Dwaipayan Shikari last updated on 17/Jul/20
	$3)∫((cosu)/(1+sin^2 u))du=∫(dt/(1+t^2 ))=tan^(−1) sinu+C { put sinu=t$
	$3) \int \frac{c o s u}{1 + {s i n}^{2} u} d u = \int \frac{d t}{1 + t^{2}} = {t a n}^{- 1} s i n u + C {p u t s i n u = t$
	Answered by Dwaipayan Shikari last updated on 17/Jul/20

	$8) \int {s i n}^{2} {u c o s}^{4} u d u$ $\frac{1}{8} \int {s i n}^{2} 2 u (1 + c o s 2 u)$ $\frac{1}{16} \int (1 - c o s 4 u) (1 + c o s 2 u)$ $= \frac{u}{16} - \frac{s i n 4 u}{64} + \frac{s i n 2 u}{32} - \frac{1}{32} \int c o s 6 u + c o s 2 u$ $= \frac{u}{16} - \frac{s i n 4 u}{16} + \frac{s i n 2 u}{32} - \frac{s i n 6 u}{192} - \frac{c o s 2 u}{64} + C$
	Answered by Dwaipayan Shikari last updated on 17/Jul/20
	$5)∫(du/(sinu+tanu))=∫((cosu)/(sinu(cosu+1)))=2∫(((1−t^2 )/(1+t^2 ))/(((2t)/(1+t^2 ))((2/(1+t^2 ))))).(1/(1+t^2 ))=∫((1−t^2 )/(2t)) =(1/2)logt−(t^2 /4)+C=(1/2)log(tan(u/2))−(((tan(u/2))^2 )/4)+C {put t=tan(u/2)$
	$5) \int \frac{d u}{s i n u + t a n u} = \int \frac{c o s u}{s i n u (c o s u + 1)} = 2 \int \frac{\frac{1 - t^{2}}{1 + t^{2}}}{\frac{2 t}{1 + t^{2}} (\frac{2}{1 + t^{2}})} . \frac{1}{1 + t^{2}} = \int \frac{1 - t^{2}}{2 t}$ $= \frac{1}{2} l o g t - \frac{t^{2}}{4} + C = \frac{1}{2} l o g (t a n \frac{u}{2}) - \frac{{(t a n \frac{u}{2})}^{2}}{4} + C {p u t t = t a n \frac{u}{2}$
	Answered by mathmax by abdo last updated on 17/Jul/20

	$1) I = \int \frac{2 udu}{\sqrt{16 - u^{4}}} changement u^{2} = x give 2 udu = dx \Rightarrow$ $I = \int \frac{dx}{\sqrt{16 - x^{2}}} =_{x = 4 t} \int \frac{4 dt}{4 \sqrt{1 - t^{2}}} = arcsint + C$ $= \arcsin (\frac{x}{4}) + c = \arcsin (\frac{u^{2}}{4}) + c .$
	Answered by mathmax by abdo last updated on 17/Jul/20

	$2) we do the changement x = \cos (2 t) \Rightarrow 2 t = arcosx and$ $\int \sqrt{\frac{1 + x}{1 - x}} dx = \int \sqrt{\frac{1 + \cos (2 t)}{1 - \cos (2 t)}} (- 2 \sin (2 t) dt$ $= - 2 \int \sqrt{\frac{{2 \cos}^{2} t}{{2 \sin}^{2} t}} \sin (2 t) dt = - 2 \int \frac{cost}{sint} \times 2 sint cost dt$ $= - 4 \int \cos^{2} t dt = - 4 \int \frac{1 + \cos (2 t)}{2} dt = - 2 \int (1 + \cos (2 t)) dt$ $= - 2 t - \sin (2 t) + C = - arcosx - \sqrt{1 - x^{2}} + C$
	Answered by bemath last updated on 18/Jul/20

	$(7) \int \frac{\sqrt{1 - \cos u}}{2} d u = \int \frac{\sqrt{1 - (1 - 2 \sin^{2} (\frac{u}{2})}}{2} d u$ $= \frac{\sqrt{2}}{2} \int \sin (\frac{u}{2}) d u$ $= - \sqrt{2} \cos (\frac{u}{2}) + C$
	Answered by mathmax by abdo last updated on 18/Jul/20
	$I =∫ sin^2 x cos^4 xdx ⇒I =∫ sin^2 x cos^2 x cos^2 xdx =∫ (((sin(2x))/2))^2 cos^2 x dx =(1/4) ∫ sin^2 (2x)cos^2 x dx =(1/4) ∫(((1−cos(4x))/2))(((1+cos(2x))/2))dx =(1/(16)) ∫(1+cos(2x)−cos(4x)−cos(2x)cos(4x))dx =(1/(16)){ x+(1/2)sin(2x)−(1/4)sin(4x)}−(1/(16)) ∫cos(2x)cos(4x)dx ∫cos(2x)cos(4x)dx =(1/2)∫(cos(6x)+cos(2x))dx =(1/(12))sin(6x)+(1/4)sin(2x) ⇒ I =(x/(16)) +((sin(2x))/(32)) −((sin(4x))/(64)) −(1/(16.12))sin(6x)−((sin(2x))/(64)) +C =(x/(16)) +((sin(2x))/(64)) −((sin(4x))/(64))−((sin(6x))/(192)) +C$
	$I = \int \sin^{2} x \cos^{4} xdx \Rightarrow I = \int \sin^{2} x \cos^{2} x \cos^{2} xdx$ $= \int {(\frac{\sin (2 x)}{2})}^{2} \cos^{2} x dx = \frac{1}{4} \int \sin^{2} (2 x) \cos^{2} x dx$ $= \frac{1}{4} \int (\frac{1 - \cos (4 x)}{2}) (\frac{1 + \cos (2 x)}{2}) dx$ $= \frac{1}{16} \int (1 + \cos (2 x) - \cos (4 x) - \cos (2 x) \cos (4 x)) dx$ $= \frac{1}{16} {x + \frac{1}{2} \sin (2 x) - \frac{1}{4} \sin (4 x)} - \frac{1}{16} \int \cos (2 x) \cos (4 x) dx$ $\int \cos (2 x) \cos (4 x) dx = \frac{1}{2} \int (\cos (6 x) + \cos (2 x)) dx$ $= \frac{1}{12} \sin (6 x) + \frac{1}{4} \sin (2 x) \Rightarrow$ $I = \frac{x}{16} + \frac{\sin (2 x)}{32} - \frac{\sin (4 x)}{64} - \frac{1}{16.12} \sin (6 x) - \frac{\sin (2 x)}{64} + C$ $= \frac{x}{16} + \frac{\sin (2 x)}{64} - \frac{\sin (4 x)}{64} - \frac{\sin (6 x)}{192} + C$
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