Question

integration of sec(tan^-1 x)

Answer 1

Hi there :)

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inverse trig huh? nyc.....    ;)

now , coming to the question......

first of all, for these type of questions.... imagine a triangle(right angled) .....

and assuming you already know all the ratios... i.e;

sin(x)= opp/hyp                            cos(x)= adj/hyp

tan(x)= sin(x)/cos(x)= opp/adj     cot(x) = 1/tan

sec(x)= 1/cos(x)                             cosec(x)= 1/sin

taking  (x) as an angle of the triangle and taking ratios wrt (x)....

               .

               | \

  (opp)    |    \   (hyp)

               |      \

               |___x\

                 (adj)

(probably you know all this)

then the real part of the answer comes here....... >

               

any inverse funtion is an angle.... i.e; it is like (x)

and this angle can b used interchangably between the ratios like so....

1) sin^-1 x = cos^-1 (√(1-x²) )        

 (why??  well.... look at this) bcause sin^-1 x an angle, i.e; sin^-1 x =Ф

⇒ sin Ф = x/1  (just subtitute in the Δ and find all sides to get other ratios)

               .

               |  \

       x      |     \    1

               |       \

               |___Ф\

                √(1-x²)

now because cos Ф = √(1-x²) /1

we can say Ф =   cos^-1 (√(1-x²) )

and  cos^-1 (√(1-x²) ) =sin^-1 x      (as simple as that)

similarly we obtain results for all other ratios...

.........

tain^-1 x = sec^-1 (√(x²+1) / 1)

and as sec(sec^-1 Ф) = Ф

we get sec(sec^-1 (√(x²+1))) =  √(x²+1) [Ans.]

and then,...... we integrate..........  :)

$\int{\sqrt{x^{2}+1} } \, dx$

as we know the formula

$\int{\sqrt{x^{2}+a^{2} } } \, dx$

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

we use this to obtain answer as

$\int{\sqrt{x^{2}+1} } \, dx$

= $\frac{x}{2}\sqrt{x^{2}+1}+\frac{1}{2}sinh^{-1}(x)+c$

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Cheers : )