If `inte^(x)((1-sinx)/(1-cosx))dx=f(x)+` Constant, then f(x) is equal to

Options

A. `e^(x)cot((x)/(2))+C`
B. `e^(-x)cot((x)/(2))+C`
C. `-e^(x)cot((x)/(2))+C`
D. `-e^(-x)cot ((x)/(2))+C`

Answer

C

Explanation

We have ,
`I=inte^(x)((1-sinx)/(1-cosx))dx`
`rArrI=inte^(x)((1)/(2)"cosec"^(2)(x)/(2)-"cot"(x)/(2))dx`
`rArr I=inte^(x)(underset(f)(-cot)(x)/(2)+(1)/(2)underset(f')("cosec"^(2))(x)/(2))dx=-e^(x)"cot"(x)/(2)+C`
THEOREM Prove that
(i) `int^(ax)sinbxdx=(e^(ax))/(a^(2)+b^(2))(a sin bx-b cosbx)+C`
(ii) `inte^(ax)cos bx dx = (e^(ax))/(a^(2)+b^(2))(a cos bx -b sin bx)+C`
PROOF (i) Let `I=inte^(ax)sinbx dx`. then ,
`I=intunderset(I)(e^(ax))underset(II)(sin) bx dx=-e^(ax)sinbx dx =e^(ax)(cos bx)/(b)-intae^(ax)((-cosbx)/(b))dx`
`rArr I=-(1)/(b)e^(ax)cosbx +(a)/(b)intunderset(I)(e^(ax))underset(II)(cos) " bx dx"`
`rArrI=-(1)/(b)e^(ax)cosbx+(a)/(b)[e^(ax)(sin bx)/(b)-intae^(ax)(sinbx)/(b)dx]`
`rArrI=-(1)/(b)e^(ax)cosbx +(a)/(b^(2))e^(ax)sinbx-(a^(2))/(b^(2))inte^(ax)sin " bx dx"`
`rArr I=-(1)/(b)e^(ax)cosbx +(a)/(b^(2))e^(ax)sinbx-(a^(2))/(b^(2))I`.
`:. I+(a^(2))/(b^(2))I=(e^(ax))/(b^(2))(a sinbx - b cos bx)`
`rArr I((a^(2)+b^(2))/(b^(2)))=(e^(ax))/(b^(2))(a sin - b cos bx)`
Or , `=(e^(ax))/(a^(2)+b^(2))(asin bx - b cos bx)+C`.
(ii) Proceed as in (i).