- #1

- 493

- 8

My work...

integral of sinx cox dx

= 1/2 integral of 2 sinx cos dx

= 1/2 integral of sin 2x dx

u = 2x

du = 2

so 1/2 * 1/2 of integral of sin u du

= 1/4 [-cos u] + c

= - 1/4 cos 2x + c

**is this correct?**

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- Thread starter Natasha1
- Start date

- #1

- 493

- 8

My work...

integral of sinx cox dx

= 1/2 integral of 2 sinx cos dx

= 1/2 integral of sin 2x dx

u = 2x

du = 2

so 1/2 * 1/2 of integral of sin u du

= 1/4 [-cos u] + c

= - 1/4 cos 2x + c

- #2

arildno

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Quite so!

- #3

0rthodontist

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Yes, it is, except you should say du = 2 dx instead of just 2.

- #4

- 493

- 8

0rthodontist said:Yes, it is, except you should say du = 2 dx instead of just 2.

Great. So if we know look at the following integral x sin x cos x dx

My work....

Let u = x

du/dx = 1

dv/dx = sin x cos x

v = -1/4 cos 2x (from above)

so = -1/4 x cos x - integral of -1/4 cos 2x dx

= -1/4 x cos 2x + 1/4 integral of cos 2x dx

Let u = 2x

du/dx = 2

so = -1/4 x cos 2x + 1/4 * 1/2 integral of cos u du

= -1/4 x cos 2x + 1/8 sin 2x + c

- #5

arildno

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Yes it is, as can be verified by differentiating your expression.

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