Question

What is the antiderivative of `1/sinx`?

Answer 1

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.

Answer 2

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`