How can this integral be solved? int[0,1] cos(2x + 5) dx

khansaheb said:
If I were to solve this problem
I would start with substitution

u = 2*x +5

I would also convert dx and the "integration limits" accordingly.

Please show us what you have tried and exactly where you are stuck.

Please follow the rules of posting in this forum, as enunciated at:


Please share your work/thoughts about this problem
Click to expand...
Alright,
and du = 2.

What do we do after that?
 
du = 2 dx

substitute and reset the limits of integration from x to u values
 
lasvegas666 said:
Okey doke.

Are we getting closer to the final answer?
Click to expand...

I would like to know that you have covered similar problems in a course and have notes on it. You have been
asking for little steps along the way very frequently instead of writing down in the thread much progress in
one post. As tutors/helpers, we are here to help reinforce concepts with what you are already supposed to have
learned, not teach you how to solve the problem, and especially not do it with you not being a participant by
not sharing your work in the posts.
 
willbill said:
change the dx to d(2x+5)
Click to expand...
That would be incorrect.

lasvegas666 said:
Evaluate: Integrate [0 to 1] cos(2x+5) dx
Click to expand...


\(\displaystyle \int_{0}^{1}cos(2x+5) dx\).....................substitute

u = (2x+5).....................&

du = 2 dx

Change Upper limit of integration.........1 \(\displaystyle \to\) 7................and

Change Lower limit of integration.........0 \(\displaystyle \to\) 5................thus

\(\displaystyle \int_{0}^{1}cos(2x+5) dx\) ................. transforms to

\(\displaystyle \int_{5}^{7}\frac{cos(u)}{2} du\)

=(1/2) * [sin(7) - sin(5)]..............Now use your calculator to finish it
 
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willbill said:
I would change the [math]dx[/math] to [math]d(2x + 5)[/math]
Click to expand...
You mean:

[imath]\displaystyle dx = \frac{1}{2}d(2x + 5)[/imath]

Then

[imath]\displaystyle \frac{1}{2}\int_{0}^{1} \cos(2x + 5) \ d(2x + 5) = \frac{1}{2}\sin(2x + 5)\bigg|_0^1[/imath]
 
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