IIT JEE , AIEEE Forum - Mathematics , Integral Calculus - integerate the following please 1/cosx-sinx .dx (dont use the partial fraction)

asmit

cosx-sinx can be written as (2)^.5{cos45cosx-sin45sinx)

=(2)^.5{cos(x+45)}.Now cos(x+45) is in denominator and in numerator it becomes sec function.There is a standard form for integrating secx.Consult your text book.

swashata4iit

_{[ ]}

^{[ ]}dx/(cosx-sinx)
= 1/

2

1/(cosxcos

/2-sinxsin

/2)dx
=1/

2

sec(x+

/2)dx
=1/

2log[sec(x+

/2)+tan(x+

/2)] +c

See its very simple
for u here i mention some formula

_{[ ]}

^{[ ]}secxdx=log[secx+tanx] +c
_{[ ]}

^{[ ]} cosecxdx = log[cosecx-cotx] + c

swashata4iit

PLEASE VOTE ME IF USEFUL

amar.gupta

good work asmit.

Dear swashata4iit,

it will be pie/4 not pie/2, rest approach is fine

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