What is the antiderivative of #1/sinx#?

2 Answers
Jim H
Sep 19, 2016

It is #-ln abs(cscx + cot x)#

Explanation:

#1/sinx = cscx = cscx (cscx+cotx)/(cscx+cotx)#

# = (csc^2 x + csc x cot x)/(cscx+cotx)#

The numerator is the opposite (the 'negative') of the derivative of the denomoinator.

So the antiderivative is minus the natural logarithm of the denominator.

#-ln abs(cscx + cot x)#.

(If you've learned the technique of substitution, we can use #u = cscx + cot x#, so #du = -csc^2 x - cscx cotx#. The expression becomes #-1/u du#.)

You can verify this answer by differentiating.

Jim S
May 7, 2018

A different approach to it

#int1/sinxdx# #=#

#intsinx/sin^2xdx#

#intsinx/(1-cos^2x)dx#

Substitute

#cosx=u#

#-sinxdx=du#

#sinxdx=-du#

#=# #-int1/(1-u^2)du#

  • #1/(1-u^2)=1/((u-1)(u+1))=A/(u-1)+B/(u+1)# #=#

#(A(u+1)+B(u-1))/((u-1)(u+1))#

We need #A(u+1)+B(u-1)=1# #<=>#

#Au+A+Bu-B=1# #<=>#

#(A+B)u+A-B=1# #<=>#

#(A+B)u+A-B=0u+1# #<=>#

#{(A+B=0" "),(A-B=1" "):}# #<=>#

#{(A+B=0" "),(A=B+1" "):}# #<=>#

#{(B+1+B=0" "),(A=B+1" "):}# #<=>#

#{(B=-1/2" "),(A=1/2" "):}#

Therefore, #-int1/(1-u^2)du# #=#

#-int((1/2)/(u-1)-(1/2)/(u+1))du# #=#

#1/2int(1/(u+1)-1/(u-1))du# #=#

#1/2int(((u+1)')/(u+1)-((u-1)')/(u-1))du# #=#

#1/2(ln|u+1|-ln|u-1|+c)# #=#

#1/2(ln|(u+1)/(u-1)|+c)# #=#

#1/2(ln|(cosx+1)/(cosx-1)|+c)# #=#

#1/2(ln|(1-cosx)/(1+cosx)|+c)#

#ln|tan(x/2)|+c'# ,

#(c,c')##in##RR#

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