Math, asked by RoshanSingh1, 1 year ago

cos(sin inverse x) = 1/9
solve it

RoshanSingh1: but not i

RoshanSingh1: can u solve integration bypart

AnshikaGautam: yess

RoshanSingh1: okkk

RoshanSingh1: intregrals of [x. log(x plus 1 )dx]

AnshikaGautam: x^2/2.log(x+1)+x.1/(x+1)

AnshikaGautam: Sorry it will be

AnshikaGautam: [x^2log(x+1)]/2- {x^2/2-x+log(x+1)}/2ln10

RoshanSingh1: your language is very difficult to understand

RoshanSingh1: but I m trying to get

Answers

Answered by kaustuvreigns

Here is the solution... Ahhh yessiirr!!

Attachments:

RoshanSingh1: correct......... friend

Answered by xantusarcar18

At first we simplify for Cos(Sin-1 x) -

Let, Sin-1X = y, then X = Sin y

Now putting the value of X in Cos(sin-1x), we get -

Cos{sin-1(siny)} = cos y ----- (i)

Again, we know that,

Sin2y + Cos2y = 1

or Cos2y = 1 – sin2y

or Cos y = ( 1 – sin2y)1/2 ----------- (II)

From (i) and (II),

Cos{sin-1(siny)} = (1-sin2y)1/2

or, Cos (sin-1 x) = ( 1 – x2)1/2

Now, cos(sin-1 x) = 1/9

or, ( 1 – x2)1/2 =1/9

or, (1-x2) = 1/ 81

or, X2 = 1 – 1/81

or, X2 = 80/81

or, x = ? ( solve it)

Previous Question

Next Question