Integral of Sin: why does dv= sin(1/4 x)dx equal v= -4cos(1/4 x) + C ?

[BigNate]

Hello Everyone,

I can't comprehend what I feel like should be an easy question. I know that integral of sinx dx = -cosx +C. So if I have the following: dv= sin(1/4 x)dx, why is the integral v= -4cos(1/4 x) + C and not just -cos(1/4x)+C?

In other words, where does the -4 come from? Can someone please explain that to me?

Thanks for your time!

 

[Deleted member 4993]

Hello Everyone,

I can't comprehend what I feel like should be an easy question. I know that integral of sinx dx = -cosx +C. So if I have the following: dv= sin(1/4 x)dx, why is the integral v= -4cos(1/4 x) + C and not just -cos(1/4x)+C?

In other words, where does the -4 come from? Can someone please explain that to me?

Thanks for your time!

Differentiate:

y = -cos(x/4) + C ......... don't forget chain rule....

What do you get?

Now differentiate:

y = -4 * cos(x/4) + C

 

[BigNate]

Ahhh, I see. Thank you!

 

[Steven G]

Hello Everyone,

I can't comprehend what I feel like should be an easy question. I know that integral of sinx dx = -cosx +C. So if I have the following: dv= sin(1/4 x)dx, why is the integral v= -4cos(1/4 x) + C and not just -cos(1/4x)+C?

In other words, where does the -4 come from? Can someone please explain that to me?

Thanks for your time!

Subhotosh gave the reason why your answer is wrong. So you ask so how do we get the correct answer.

I ALWAYS let u = the angle, unless the angle is x, y, z,...

\(\displaystyle if \ u \ = \dfrac{1}{4}x, \ then \ du \ = \dfrac{1}{4}dx \)

Now substitute this into your integral and show us what you get.