IIT JEE , AIEEE Forum - Mathematics , Integral Calculus

kislay

evakuate

₀

^10pi ([arcsecx] +[arccotx])dx,where[.] denotes greatest integer function
ans...
10pi-sec1

elastiboysai

_{[1 ]}

^{[10pi ]} [arc secx]+[arc cotx] dx

_{[1 ]}

^{[sec1 ]} [arc secx]+[arc cotx] dx+ _{[ sec1]}

^{[10pi ]}[arc secx]+[arc cotx] dx

_{[ 1]}

^{[sec1 ]} (0+ 0) +_{[ sec1]}^{[10pi ] (1+0) dx}

^{gives 10pi - sec1}

^{n chec d qn.}

^{sec inverse 0 - 1 is obv not defined}

sboosy

I think the question shud have been 1 to 10 pi and not o to 10 pi

bcos sec inverse not defined for values between -1 to 1

I think this is what is elastic has assumed

anyway good work

magiclko

the graph for the given function is as shown in attachd figure,

blue is arcsec x, yellow is [acsec x], green is arccot x, and red is for [arccot x] and the area blackend will be the answer for given limits...

kislay

as evident from the graph given by maigclco
ans.shoul also include cot1....
ie final ans sud be........
10pi-sec1+cot1.......
i had also got the same.

magiclko

then kislay u r absolutely right

.... have confidence with the answers u get...

Greatdreams

I am no big expert in graphs, however I still can't understand where elastiboysai went wrong!!
Please some one explain.

kislay

it will b clear if we solve the two given integrals [arcsecx] and [arccotx] seperately .......ans will come 10pi-sec1+cot1.......

elastiboysai

Hi,

It's a problem wid d question

obviously

sec in verse is not defined between 0 - 1

so i have taken the question to be 1- 10pi

bt kislay has neglected secin verse in dat interval and has taken cot inverse's integral.. so the confus..

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