∫(dx/(√((a+1)cos 2x +4cos x −a+3)))=?


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Question Number 49838 by MJS last updated on 11/Dec/18

$\int \frac{d x}{\sqrt{(a + 1) \cos 2 x + 4 \cos x - a + 3}} = ?$
Commented by MJS last updated on 11/Dec/18

$\int \frac{d x}{\sqrt{(a + 1) \cos 2 x + 4 \cos x - a + 3}} =$ $[t = \frac{x}{2} \to d x = 2 d t]$ $= 2 \int \frac{d t}{\sqrt{(a + 1) \cos 4 t + 4 \cos 2 t - a + 3}} =$ $= \frac{\sqrt{2}}{2} \int \frac{d t}{\cos t \sqrt{(a + 1) \cos^{2} t - a}} =$ $[u = \tan t \to d t = \frac{d u}{u^{2} + 1}; \cos t = \frac{1}{\sqrt{u^{2} + 1}}]$ $= \frac{\sqrt{2}}{2} \int \frac{d u}{\sqrt{1 - {a u}^{2}}} =$ $[v = \sqrt{a} u \to d u = \frac{d v}{\sqrt{a}}]$ $= \frac{1}{\sqrt{2 a}} \int \frac{d v}{\sqrt{1 - v^{2}}} = \frac{1}{\sqrt{2 a}} \arcsin v = \frac{1}{\sqrt{2 a}} \arcsin (\sqrt{a} u) =$ $= \frac{1}{\sqrt{2 a}} \arcsin (\sqrt{a} \tan t) = \frac{1}{\sqrt{2 a}} \arcsin (\sqrt{a} \tan \frac{x}{2}) + C$
	Answered by tanmay.chaudhury50@gmail.com last updated on 11/Dec/18
	$∫(dx/(√(−a(1−cos2x)+cos2x+4cosx+3))) ∫(dx/(√(4cosx+4−(1−cos2x)−a(1−cos2x)))) ∫(dx/(√(4(2cos^2 (x/2))−(2sin^2 x)−a(2sin^2 x)))) ∫(dx/(√(8cos^2 (x/2)−2(a+1)×4sin^2 (x/2)cos^2 (x/2)))) ∫((sec(x/2)dx)/(√(8−8(a+1)sin^2 (x/2)))) (1/(√8))×∫((sec(x/2)dx)/(√(1−(a+1)sin^2 (x/2)))) (1/((√(8(a+1))) ))∫((sec(x/2)dx)/(√((1/(a+1))−sin^2 (x/2)))) (1/(√(8(a+1))))∫((cos(x/2)dx)/(cos^2 (x/2)(√((1/(a+1))−sin^2 (x/2)))))dx t=sin(x/2) dt=(1/2)cos(x/2)dx (1/(√(8(a+1))))∫((2dt)/((1−t^2 )(√((1/(a+1))−t^2 )))) t=(1/k) dt=((−1)/k^2 )dk (1/(√(2(a+1))))∫((−dk)/(k^2 (1−(1/k^2 ))(√((1/(a+1))−(1/k^2 ))))) (1/(√(2(a+1))))∫((−kdk)/((k^2 −1)(√((k^2 /(a+1))−1)))) ((−1)/(√(2(a+1))))∫((kdk)/((k^2 −1)(√((k^2 /(a+1))−1)))) p=k^2 dp=2kdk ((−1)/(√(2(a+1))))∫(dp/(2(p−1)(√((p/(a+1))−1)))) ((−1)/((√(8(a+1))) ))∫(dp/((p−1)(√((p/(a+1))−1)))) y^2 =(p/(a+1))−1 2ydy=(dp/(a+1)) =((−1)/(√(8(a+1))))∫((2(a+1)ydy)/({(a+1)(y^2 +1)−1}×y)) =((−2(a+1))/(√(8(a+1))))∫(dy/({(a+1)y^2 +a})) =((−((√(a+1)) ))/(√2))×(1/(a+1))∫(dy/(y^2 +(a/(a+1)))) =((−1)/(√(2(a+1))))×(1/(√(a/(a+1))))tan^(−1) ((y/(√(a/(a+1)))))+c ((−1)/(√(2a)))tan^(−1) ((((√((p/(a+1))−1)) )/(√(a/(a+1)))))+c ((−1)/(√(2a)))tan^(−1) (((√((k^2 /(a+1))−1))/(√(a/(a+1))))) =((−1)/(√(2a)))tan^(−1) ((((√((1/(t^2 (a+1)))−1)) )/(√(a/(a+1))))) =((−1)/(√(2a)))tan^(−1) (((√((1/(sin^2 ((x/2))(a+1)))−1))/(√(a/(a+1)))))+c$
	$\int \frac{d x}{\sqrt{- a (1 - c o s 2 x) + c o s 2 x + 4 c o s x + 3}}$ $\int \frac{d x}{\sqrt{4 c o s x + 4 - (1 - c o s 2 x) - a (1 - c o s 2 x)}}$ $\int \frac{d x}{\sqrt{4 (2 {c o s}^{2} \frac{x}{2}) - (2 {s i n}^{2} x) - a (2 {s i n}^{2} x)}}$ $\int \frac{d x}{\sqrt{8 {c o s}^{2} \frac{x}{2} - 2 (a + 1) \times 4 {s i n}^{2} \frac{x}{2} {c o s}^{2} \frac{x}{2}}}$ $\int \frac{s e c \frac{x}{2} d x}{\sqrt{8 - 8 (a + 1) {s i n}^{2} \frac{x}{2}}}$ $\frac{1}{\sqrt{8}} \times \int \frac{s e c \frac{x}{2} d x}{\sqrt{1 - (a + 1) {s i n}^{2} \frac{x}{2}}}$ $\frac{1}{\sqrt{8 (a + 1)}} \int \frac{s e c \frac{x}{2} d x}{\sqrt{\frac{1}{a + 1} - {s i n}^{2} \frac{x}{2}}}$ $\frac{1}{\sqrt{8 (a + 1)}} \int \frac{c o s \frac{x}{2} d x}{{c o s}^{2} \frac{x}{2} \sqrt{\frac{1}{a + 1} - {s i n}^{2} \frac{x}{2}}} d x$ $t = s i n \frac{x}{2} d t = \frac{1}{2} c o s \frac{x}{2} d x$ $\frac{1}{\sqrt{8 (a + 1)}} \int \frac{2 d t}{(1 - t^{2}) \sqrt{\frac{1}{a + 1} - t^{2}}}$ $t = \frac{1}{k} d t = \frac{- 1}{k^{2}} d k$ $\frac{1}{\sqrt{2 (a + 1)}} \int \frac{- d k}{k^{2} (1 - \frac{1}{k^{2}}) \sqrt{\frac{1}{a + 1} - \frac{1}{k^{2}}}}$ $\frac{1}{\sqrt{2 (a + 1)}} \int \frac{- k d k}{(k^{2} - 1) \sqrt{\frac{k^{2}}{a + 1} - 1}}$ $\frac{- 1}{\sqrt{2 (a + 1)}} \int \frac{k d k}{(k^{2} - 1) \sqrt{\frac{k^{2}}{a + 1} - 1}}$ $p = k^{2} d p = 2 k d k$ $\frac{- 1}{\sqrt{2 (a + 1)}} \int \frac{d p}{2 (p - 1) \sqrt{\frac{p}{a + 1} - 1}}$ $\frac{- 1}{\sqrt{8 (a + 1)}} \int \frac{d p}{(p - 1) \sqrt{\frac{p}{a + 1} - 1}}$ $y^{2} = \frac{p}{a + 1} - 1 2 y d y = \frac{d p}{a + 1}$ $= \frac{- 1}{\sqrt{8 (a + 1)}} \int \frac{2 (a + 1) y d y}{{(a + 1) (y^{2} + 1) - 1} \times y}$ $= \frac{- 2 (a + 1)}{\sqrt{8 (a + 1)}} \int \frac{d y}{{(a + 1) y^{2} + a}}$ $= \frac{- (\sqrt{a + 1})}{\sqrt{2}} \times \frac{1}{a + 1} \int \frac{d y}{y^{2} + \frac{a}{a + 1}}$ $= \frac{- 1}{\sqrt{2 (a + 1)}} \times \frac{1}{\sqrt{\frac{a}{a + 1}}} {t a n}^{- 1} (\frac{y}{\sqrt{\frac{a}{a + 1}}}) + c$ $\frac{- 1}{\sqrt{2 a}} {t a n}^{- 1} (\frac{\sqrt{\frac{p}{a + 1} - 1}}{\sqrt{\frac{a}{a + 1}}}) + c$ $\frac{- 1}{\sqrt{2 a}} {t a n}^{- 1} (\frac{\sqrt{\frac{k^{2}}{a + 1} - 1}}{\sqrt{\frac{a}{a + 1}}})$ $= \frac{- 1}{\sqrt{2 a}} {t a n}^{- 1} (\frac{\sqrt{\frac{1}{t^{2} (a + 1)} - 1}}{\sqrt{\frac{a}{a + 1}}})$ $= \frac{- 1}{\sqrt{2 a}} {t a n}^{- 1} (\frac{\sqrt{\frac{1}{{s i n}^{2} (\frac{x}{2}) (a + 1)} - 1}}{\sqrt{\frac{a}{a + 1}}}) + c$
	Commented by MJS last updated on 11/Dec/18

	$correct, thank you!$
	Commented by malwaan last updated on 12/Dec/18

	$great$ $thank you$
	Commented by tanmay.chaudhury50@gmail.com last updated on 12/Dec/18

	$M J S s i r y o u r m e t h o d i s s h o r t a n d t r i c k y,$ $t h a n k y o u s i r . . .$
	Commented by MJS last updated on 12/Dec/18

	${you}^{'} re welcome . I^{'} m an integral addicted . . .$
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