Antiderivatives Part IV

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Hckyplayer8

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Find the indefinite integral of (2x - 3 cos2x)dx

f'(x2) = 2x

I broke out the product and chain rule and started plugging in values I knew.

f was harder so I let that go till last.

g'(x) = cos2x

g(x) = sin2x

So now of the product rule I have f(x) * cos2x + f'(x) * sin 2x

f'(x) should probably be a 0 so that means f is a constant.

So now based on the chain rule (The second portion of the chain rule is the derivative of 2x = 2), I need to figure out what times 2 equals -3.

f(x)=-3/2

Thus the answer to the problem is x2 - 3/2 sin2x

Are there any mistakes?
 
Last edited:
Hckyplayer8 said:
Find the indefinite integral of (2x - 3 cos2x)dx

f'(x2) = 2x

I broke out the product and chain rule and started plugging in values I knew.

f was harder so I let that go till last.

g'(x) = cos2x

g(x) = sin2x

So now of the product rule I have f(x) * cos2x + f'(x) * sin 2x

f'(x) should probably be a 0 so that means f is a constant.

So now based on the chain rule (The second portion of the chain rule is the derivative of 2x = 2), I need to figure out what times 2 equals -3.

f(x)=-3/2

Thus the answer to the problem is x2 - 3/2 sin2x

Are there any mistakes?
Click to expand...
Did you calculate the derivative of your answer? That will tell you if there are mistakes.
 
Subhotosh Khan said:
Did you calculate the derivative of your answer? That will tell you if there are mistakes.
Click to expand...

I understand the concept of double checking the work to make sure the derivative and antiderivative are complementary to one another.

Going back through it looks good to me but then again I'll call a spade a spade. I'm a C to maybe a B- student in this material which means I'm prone to errors in even the by now well practiced derivative portion of the answer check.

I've checked my work in Symbolab but I also know those things are fickle and can lie so...
 
Hckyplayer8 said:
I understand the concept of double checking the work to make sure the derivative and antiderivative are complementary to one another.

Going back through it looks good to me but then again I'll call a spade a spade. I'm a C to maybe a B- student in this material which means I'm prone to errors in even the by now well practiced derivative portion of the answer check.

I've checked my work in Symbolab but I also know those things are fickle and can lie so...
Click to expand...
But:

Did you calculate the derivative of your answer? What did you get?
 
Subhotosh Khan said:
But:

Did you calculate the derivative of your answer? What did you get?
Click to expand...

I did and I got the original expression. So I assume it's correct but based on how I thought similar in my other thread...it could be bad math on my part with a magic answer confirming a false positive...if you will.
 
Please note the following about the notation cos2x.
cos(2x) is the cosine of twice the angle x.

cos^2(x) = cos(x) cos(x) = (cos(x))^2 is the square of cos(x).

Please use the parenthesis in the future!

-Dan
 
Hckyplayer8 said:
Find the indefinite integral of (2x - 3 cos2x)dx

f'(x2) = 2x

I broke out the product and chain rule and started plugging in values I knew.

f was harder so I let that go till last.

g'(x) = cos2x

g(x) = sin2x

So now of the product rule I have f(x) * cos2x + f'(x) * sin 2x

f'(x) should probably be a 0 so that means f is a constant.

So now based on the chain rule (The second portion of the chain rule is the derivative of 2x = 2), I need to figure out what times 2 equals -3.

f(x)=-3/2

Thus the answer to the problem is x2 - 3/2 sin2x

Are there any mistakes?
Click to expand...
Can you explain how you know that f'(x2) = 2x?.
Do you not want to use u-substitution to solve integrals? Just wondering?
 
You ONLY know how to compute the int(cos x)dx or int(cos u)du or ... or int(cos q)dq. You had int(cos(2x))dx
If I were you, for all integrals that involve trig functions I would let u = angle unless the angle is simply x or u or y or q etc.
 
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