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In solving the integration of: (sin x) / (1+ cos x)  [0, pi/2] where u= (1+ cos x) and du = -sin x dx, the solution converted the interval of [0, pi/2] to [2, 1].       I don't understand why this was necessary since the order of the intervals seem reversed. Could you give me a step by step explanation why this was necessary?

Question

In solving the integration of: (sin x) / (1+ cos x)  [0, pi/2] where u= (1+ cos x) and du = -sin x dx, the solution converted the interval of [0, pi/2] to [2, 1].       I don't understand why this was necessary since the order of the intervals seem reversed. Could you give me a step by step explanation why this was necessary?

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Step 1

The function is

si1integrated on the interval
1cosx
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si1integrated on the interval 1cosx

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Step 2

To compute:

я
sin x
d
0 1+cosX
Take 1 cosx= t; differentiate it with respect to t.
d
(1cosx)
dt
dt
dt
dx
1
-sin x
dt
-sin xdx dt
sin xdx dt
At x 0, the value of t becomes t =1+ cos(0)=1+1= 2
п
the value oft becomes t = 1 +cos]
2
At x -
=1+0 1
-
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я sin x d 0 1+cosX Take 1 cosx= t; differentiate it with respect to t. d (1cosx) dt dt dt dx 1 -sin x dt -sin xdx dt sin xdx dt At x 0, the value of t becomes t =1+ cos(0)=1+1= 2 п the value oft becomes t = 1 +cos] 2 At x - =1+0 1 -

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Step 3

Thus, the definite integral with...

sinx
- dx
1cosx
0
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sinx - dx 1cosx 0

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Tagged in

Math

Calculus

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