IIT JEE , AIEEE Forum - Mathematics , Integral Calculus

bubunbhatt

_{[ -2 ]}

^{[ 5 ]} cot~1(tanx) dx

Here cot~1represents cot inverse

(a) 7

/2 (b) 7

²/2 (c) 3

/2 (d)

²

Please post solution with full explanation

aysh

HAI BUBUNBHATT!!!
it's easy
HERE IT GOES...

take cot inv.(tanx) as (pi/2)-tan inv.(tanx)
then,
I= -(7(pi)^2)/2

PLEEEEEZE RATE ME..

bubunbhatt

Ya but the answer given is 7

²/2 and not -7

²/2

And keep in touch throuh ur nudgebook

bubunbhatt

Please Someone help..........................................

This is a discontinuous func. and so directly putting the limits wont work out......................

rishikesh_anshu

_{[ -2 ]}

^{[ 5 ]} cot~1(tanx) dx

cot~1(tanx) =periodic function,with period

the period of tanx

_{[ -2 ]}

^{[ 5 ]} cot~1(tanx) dx=7 *_{[ 0 ]}

^{[ ]} cot~1(tanx) dx

tanx =cot(

/2-x) for x belong(0,

/2)
=cot( 3

/2-x) for x belong(

/2,

)

so
_{[ 0 ]}

^{[ ]} cot~1(tanx) dx=_{[ 0 ]}

^{[ /2 ]} cot~1(tanx) dx + _[

/2_]

^{[ ]} cot~1(tanx) dx

=_{[ 0 ]}

^{[ /2 ]} (

/2-x) dx+_[

/2_]

^{[ ]} (3

/2-x) dx

²/2

multiplying by 7
7

²/2

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