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∫{[1-cosx]/[cosx(1+cosx)]}dx ∫{[1+sinx]/[sinx(1+cosx)]}dx ∫[sinx/sin4x]dx


  • ∫{[1-cosx]/[cosx(1+cosx)]}dx



  • ∫{[1+sinx]/[sinx(1+cosx)]}dx



  • ∫[sinx/sin4x]dx

Grade:12th Pass

2 Answers

Rinkoo Gupta
askIITians Faculty 80 Points
9 years ago

Dear Student,

Pls type one question in one link.

?(1-cosx)dx/cosx(1+cosx)

=?dx/cosx(1+cosx) - ?dx/(1+cosx)

=?{(1/cosx)-1/(1+cosx)}dx -?dx/(1+cosx)

=?dx/cosx -2?dx/(1+cosx)

=?secxdx-2?dx/2cos^(x/2)

=?ecxdx-?sec^2(x/2)dx

=logIsecx+tanxI-2tan(x/2)+C

Thanks & Regards

Rinkoo Gupta

AskIITians Faculty

Rishi Sharma
askIITians Faculty 646 Points
2 years ago
Dear Student,
Please find below the solution to your problem.

(A)
I=integ. of (1-cos x).dx/cos x.(1+cos x)
I=integ. of (1–1+2sin^2 x/2).dx/cos x.(1+2.cos^2 x/2 -1).
I=integ. of ( tan^2 x/2).dx/ cos x
I= integ.of (tan^2 x/2).(1+tan^2 x/2).dx /(1-tan^2 x/2).
I = integ. of (tan^2 x/2).sec^2 x/2.dx/(1-tan^2 x/2)
Let. p = tan x/2
dp = 1/2.sec^2 x/2.dx
or. 2.dp=sec ^2 x/2.dx.
I = integ.of 2.p^2.dp/(1-p^2)
I = integ.of -2(1-p^2–1).dp/(1-p^2)
I=integ.of -2[ 1. -1/(1-p^2)].dp
I= integ.of [-2 +2. 1/(1-p^2)].dp
I = -2p +2.1/2{log(p+1)-log(1-p)} + C.
I=-2.tan x/2+log(1+tan x/2)-log(1-tan x/2)+C.

(B)

645-1360_01010101.PNG

(C)

645-1887_1234421.GIF

Thanks and Regards

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