IIT JEE , AIEEE Forum - Mathematics , Integral Calculus

punnima

_{[0 ]}

^{[ /2]} [sin^-1(cosx)+cos^-1(sinx)] where [.] denotes the greatest integer function is________

(a)pi-3 (c)pi - 6

2 2

(b)3pi-9 (d) 3pi - 6

2 2

plizz explain wid complete steps (esp. where the [ ] function cums into picture as i donn understand how 2 solve probs involving box func)

hsbhatt

₀

^/2 [sin^-1(cosx) + cos^-1(sinx)] dx

Now sin^-1(cosx) + cos^-1(sinx) = (

/2-x) + (

/2-x) (because we are in the safe environs of the 1st quadrant.) =

-2x

Hence our integral is _0^/2 [pi-2x]dx and applying the substitution t =

/2-2x we get _0^/2[2t]dt. Now, the Riemann Integral for the function doesn't exist as it is not a continuous function. But I guess there is a way to tide over the points of discontinuity (and I need someone to validate this).

We can compute the integral as (leaving out the first interval where the func is zero) _0.5+

^1-dx + _{1+^{1.5 -}}2dx +_1.5+^/2 3dx = (1-

- 0.5 -

) + 2(1.5-

- 1 -

)+3(

/2 - 1.5 -

) and then let

0 giving 3

/2 - 3 which is option (d)

Now please feel free to rain bricks on my head.

punnima

well,i checked d soln..it was in y file of brill. .

it says pi-2x =3 --(1)

and hence x=(pi-3)/2

even if we use [ ] for(1) i dunno how one gets 3 ( i repeat i am not at all clear on [ ] func so i maybe rong )

hence d interval for integ was split acc ..........but i dunno how

punnima

i used d prop. _{[0 ]}

^{[b ]} f(x) dx =₀

^b f(b-x) dx

and got d integ as = 2x

is my method rite??

if not plizz temme wher im rong

hsbhatt

Kindly make your posts intelligible. like showing steps here and there. For eg you say you use the property _{[0 ]}

^{[b ]} f(x) dx =₀

^b f(b-x) dx. So you should get [2x] and not 2x. Its not clear how you arrived at 2x. and in the previous post there is an eqn pi-2x = 3. What on earth is meant by that.

kulki_1123

we have a property of definite integrals that _{[ o]}

^{[ a]} f(x)dx=_{[ 0]}

^{[a ]} f(a-x)dx

using this prop., the given integral transforms into

_{[ 0]}

^{[ ]}

/2 [arcsin(sinx)+ arccos(cosx)]dx

= _{[ 0]}

^{[ ]}

/2 [ x+x]dx

=

_{[0 ]}

^{[ ]}

/2 [2x]dx

now use the above property again...

what u ge inside the integral is..

2I =

_{[ 0]}

^{[ ]}

/2 [2x] + [

-2x]dx

now consider any two numbers (u dont need if negative numbers...not in the integral)...[a] + [b] =either [a+b} or [a+b] - 1

inside the integral if there is a 1 involved then it will get ancelled in the definite integration process

the intregral becomes..

I= _{[0 ]}

^{[ ]}

/2 [2x +

- 2x] dx

I= _{[0 ]}

^{[ ]}

/2 [

]dx

now [

] is 3

hence the answer is

3

/2

which is option (a.)

am i rite??

iitkgp_bipin

hsbhatt is correct. Good work.

kulki_1123 recheck your solution.

Once we arrived at I = _[0]

^[pi/2] [2x].dx

Split the integrals where [2x] changes its values.

Answer comes out to be 3pi/2 - 3 which is d.

hsbhatt

Thank you sir for clearing this one up.

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