Question
Asked Nov 28, 2019
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Hello, I need to determine the indefinite integral for the following:

(a). Jcosdx.
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(a). Jcosdx.

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Expert Answer

Step 1

To integrate

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Step 2

Let

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х и 2 then dx du 2 dx 2 du

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Step 3

Then,

...
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dx = Jcos (u) 2du =2cos(u du - 2sin (u)+Ccos(u) cu = sin (u) where C is the arbitrary constant

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Tagged in

Math

Calculus

Integration