Question

intregation of 1/1+cotx​

Answer 1

Answer:

(x/2) - log (sinx+cosx)+c

Step-by-step explanation:

To find---->

_______

1

∫----------------- dx

1 + cotx

Solution----->

________

let

1

I= ∫---------------dx

1+ cotx

1

=∫--------------------------dx

1 + cosx/ sinx

1

=∫-------------------------------dx

sinx +cosx/sinx

sinx

=∫-------------------------------dx

sinx +cosx

2sinx

=∫----------------------------------dx

2(sinx +cosx)

sinx +sinx

=∫---------------------------------dx

2( sinx + cosx)

(sinx+cosx)+(sinx-cosx)

=∫------------------------------------------ dx

2 ( sinx + cosx)

(sinx+ cosx)-(cosx-sinx)

=∫-------------------------------------------dx

2 (sinx+cosx)

(sinx+cosx) ( cosx -sinx)

=∫---------------------- -∫--------------------------- dx

2 ( sinx +cosx) 2 (sinx +cosx)

1 cosx -sinx

=--- ∫ 1 dx - ∫-----------------------dx

2 2( sinx +cosx)

cosx -sinx

= x/2 - ∫-----------------------dx

2( sinx+cosx)

now let

sinx+cosx =t

differentiating both sides

( cosx -sinx )dx =dt

now

dt

I=(x/2) -∫--------

t

=(x/2 ) - log t

=(x/2) - log(sinx +cosx)+c

Hope it helps you

Thanks for giving me chance to answer your question

Answer 2

here isthe solutuion,

∫ [ 1 / ( 1 + cos x/sin x ) ] dx

. = ∫ [ ( sin x ) / ( sin x + cos x ) ] dx

. = (1/2) • ∫ [ ( 2 sin x ) / ( sin x + cos x ) ] dx ...( Note This Step )

. = (1/2) • ∫ { [ ( sin x + cos x ) - ( cos x - sin x ) ] / ( sin x + cos x ) } dx ... This Too

. = (1/2) • { ∫ 1 dx - ∫ [ ( cos x - sin x ) / ( sin x + cos x ) ] dx }

. = (1/2) • { x - ∫ (1/u) du }, ... where ... u = sin x + cos x

. = ( x/2 ) - (1/2)· ln | u | + C

. = ( x/2 ) - (1/2)· ln | sin x + cos x | + C .................ans

plz approve if you are satisfied :