Exact answer to an integral with the help of the fundamental theorem

[Quantonium]

Hi!

Sorry for the vague thread name, but I'll try to explain.

I am asked to show that [MATH]\int_{0}^x cos^2(t)dt = \frac{1}{2}(sin(x)cos(x)+x)[/MATH] using the fundamental theorem.

Then I am to find an exact answer to [MATH]\int_{0}^1 \sqrt{1-x^2}dx[/MATH] using the substitution [MATH]x=sin(t)[/MATH] together with the previous problem.

I have been working with it for hours now, and can't find a solution.
I managed the first part of the problem using some trig identities, and calculating the integral in the usual manner, but the latter part has stumped me.

Any tips?

Thanks,
Mads

 

[Deleted member 4993]

Hi!

Sorry for the vague thread name, but I'll try to explain.

I am asked to show that [MATH]\int_{0}^x cos^2(t)dt = \frac{1}{2}(sin(x)cos(x)+x)[/MATH] using the fundamental theorem.

Then I am to find an exact answer to [MATH]\int_{0}^1 \sqrt{1-x^2}dx[/MATH] using the substitution [MATH]x=sin(t)[/MATH] together with the previous problem.

I have been working with it for hours now, and can't find a solution.
I managed the first part of the problem using some trig identities, and calculating the integral in the usual manner, but the latter part has stumped me.

Any tips?

Thanks,
Mads

Can you perform the integration (second part of the assignment)?

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:

https://www.freemathhelp.com/forum/threads/read-before-posting.109846/#post-486520

Please share your work/thoughts about this problem.

 

[pka]

I am asked to show that [MATH]\int_{0}^x cos^2(t)dt = \frac{1}{2}(sin(x)cos(x)+x)[/MATH] using the fundamental theorem.
Then I am to find an exact answer to [MATH]\int_{0}^1 \sqrt{1-x^2}dx[/MATH] using the substitution [MATH]x=sin(t)[/MATH] together with the previous problem. Any tips?

TIP Hint: \(\cos^2(t)=\dfrac{1}{2}\cos(2t)+\dfrac{1}{2}\)

 

[Quantonium]

Thanks for the answers.
As to the first one, I did read the rules, but decided there was nothing important for me to show. As I said, I solved the first part, and the second part didn't go further than writing the equation..

After hours of thinking I did come to an answer to the second part, though.
The key was to also change the interval [0, 1] from x-values to sin(t)-values. Giving the new interval of [0, pi/2]. After that, it was simply a matter of doing the first part again.

 

[HallsofIvy]

Hi!

Sorry for the vague thread name, but I'll try to explain.

I am asked to show that [MATH]\int_{0}^x cos^2(t)dt = \frac{1}{2}(sin(x)cos(x)+x)[/MATH] using the fundamental theorem.

No need to integrate. The "fundamental theorem of Calculus" says that [math]F(x)=\int_a^x f(t)dt[/math] if and only if [math]F'(x)= f(x)[/math] The derivative of [math]\frac{1}{2}(sin(x)cos(x)+ x)[/math] is [math]\frac{1}{2}(cos^2(x)- sin^2(x)+ 1)= \frac{1}{2}(cos^2(x)- (1- cos^2(x))+ 1)= cos^2(x)[/math].

Then I am to find an exact answer to [MATH]\int_{0}^1 \sqrt{1-x^2}dx[/MATH] using the substitution [MATH]x=sin(t)[/MATH] together with the previous problem.

Letting [math]x= sin(t)[/math], [math]dx= cos(t)dt[/math] and [math]\sqrt{1- x^2}= \sqrt{1- sin^2(t)}= cos(t)[/math]. Further, when x= 0 when t= 0 and x= 1 when [math]t= \pi/2[/math] so the integral becomes [math]\int_0^{\pi/2} cos^2(t) dt[/math]. Now use the "previous problem".

I have been working with it for hours now, and can't find a solution.
I managed the first part of the problem using some trig identities, and calculating the integral in the usual manner, but the latter part has stumped me.

Any tips?

Thanks,
Mads

 

[Steven G]

The 2nd integral also requires no integration. What curve are you trying to find the area of? What part of the curve does 0 to 1 represent?