Question

Prove That:-
$\int{\frac{dx}{1-2a\cos x+a^{2}}}\;=\;\frac{2}{1-a^2}\tan^{-1}[(\frac{1+a}{1-a})\tan\frac{x}{2}]+c$

Answer 1

Answer:hence proved✌✌

Step-by-step explanation:

=[Tex]\int{\frac{dx}{1-2acosx+a{}^{2}}[/tex]

Put cosx = $\frac{1-tan{}^{2} x/2}{ 1+tan{}^{2} x/2}}$

=[Tex]\int{\frac{(1+tan{}^{2} x/2)dx}{(1+a{}^{2} )(1+tan{}^{2} x/2)-2a(1-tan{}^{2} x/2)}}[/tex]

=[Tex]\int{\frac{(sec{}^{2} x/2)dx}{(1+a{}^{2} )(1+tan{}^{2} x/2)-2a(1-tan{}^{2} x/2)}}[/tex]

{Tan x/2 = t

Sec^2 x/2 dx = dt}

=2$\int{\frac{dt}{(1+a{}^{2})(1+t{}^{2})-2a(1-t{}^{2})}$

=2$\int{\frac{dt}{(a-1){}^{2} + t{}^{2}(a+1){}^{2}}$

=$\frac{2}{(1+a){}^{2})}$[Tex]\int{\frac{dt}{[(a-1)/(a+1)]{}^{2} +t{}^{2}}[/tex]

=$\frac{2}{(1+a)^2} \frac{ tan{}^{-1} t}{(a-1)/(a+1)}$

=-$\frac{2}{(a-1)}$$tan{}^{-1} x/2 [(1+a)/(1-a)]$ + c

=$\frac{2}{1-a}$$tan{}^{-1}(1+a)/(1-a) x/2$ + c

Hence proved

Hope it helps u

#skb✌

Answer 2

Answer:

=[Tex]\int{\frac{dx}{1-2acosx+a{}^{2}}[/tex]

Put cosx =

=[Tex]\int{\frac{(1+tan{}^{2} x/2)dx}{(1+a{}^{2} )(1+tan{}^{2} x/2)-2a(1-tan{}^{2} x/2)}}[/tex]

=[Tex]\int{\frac{(sec{}^{2} x/2)dx}{(1+a{}^{2} )(1+tan{}^{2} x/2)-2a(1-tan{}^{2} x/2)}}[/tex]

{Tan x/2 = t

Sec^2 x/2 dx = dt}

=2\int{\frac{dt}{(1+a{}^{2})(1+t{}^{2})-2a(1-t{}^{2})}

=2\int{\frac{dt}{(a-1){}^{2} + t{}^{2}(a+1){}^{2}}

=\frac{2}{(1+a){}^{2})}

(1+a)

2

)

2

[Tex]\int{\frac{dt}{[(a-1)/(a+1)]{}^{2} +t{}^{2}}[/tex]

=\frac{2}{(1+a)^2} \frac{ tan{}^{-1} t}{(a-1)/(a+1)}

(1+a)

2

2

(a−1)/(a+1)

tan

−1

t

=-\frac{2}{(a-1)}

(a−1)

2

tan{}^{-1} x/2 [(1+a)/(1-a)]tan

−1

x/2[(1+a)/(1−a)] + c

=\frac{2}{1-a}

1−a

2

tan{}^{-1}(1+a)/(1-a) x/2tan

−1

(1+a)/(1−a)x/2 + c

proved...