Haven't been taught integration by parts yet

[kamranonline]

http://usera.ImageCave.com/kamranonline/369529f005b9d646328ba8de471139.gif


Attempt to the solution. I took u=[tex]\sqrt{t}[/tex] and from there i went upto :

2[tex]\int[/tex]u[tex]^2[/tex]Sin(u^{2})dx

dunno wat to do next. Havn't been taught integration by parts yet.

 

[Unco]

http://usera.ImageCave.com/kamranonline/369529f005b9d646328ba8de471139.gif

Attempt to the solution. I took u=[tex]\sqrt{t}[/tex] and from there i went upto :

2[tex]\int[/tex]u[tex]^2[/tex]Sin(u^{2})d

dunno wat to do next. Havn't been taught integration by parts yet.

What is the question? Are you sure it isn't to find g'(x)?

 

[kamranonline]

yes it is.. sorry my bad. :P

then we have to expand the integral and and simply write the inner function by replacing t with x rite?

(i spend much time figuring out the integral, but in vain.. :P)

 

[Mark44]

http://usera.ImageCave.com/kamranonline/369529f005b9d646328ba8de471139.gif


Attempt to the solution. I took u=[tex]\sqrt{t}[/tex] and from there i went upto :

2[tex]\int[/tex]u[tex]^2[/tex]Sin(u^{2})dx

dunno wat to do next. Havn't been taught integration by parts yet.

Your substitution is not right. If u = [itex]\sqrt{t}[/itex], then du = dt/(2[itex]\sqrt{t}[/itex])
dx shouldn't even be in the integral.

 

[sutupidmath]

http://usera.ImageCave.com/kamranonline/369529f005b9d646328ba8de471139.gif


Attempt to the solution. I took u=[tex]\sqrt{t}[/tex] and from there i went upto :

2[tex]\int[/tex]u[tex]^2[/tex]Sin(u^{2})dx

dunno wat to do next. Havn't been taught integration by parts yet.

If you are interested to find g'(x), then you don't even need to do any integration at all. All, you need to do is apply 1st part of the FUndamental THeorem of Calculus. What does it say?

First manipulate your integral a little bit, like:

[tex]g(x)=-\int_{c}^{\sqrt{x}}\sqrt{t}sin(t)dt+\int_{c}^{x^2}\sqrt{t}sin(t)dt[/tex]

where c is some constant.

Now you need to directly apply FTC and the chain rule along with it and you are pretty much done.

 

[Unco]

Your substitution is not right. If u = [itex]\sqrt{t}[/itex], then du = dt/(2[itex]\sqrt{t}[/itex])
dx shouldn't even be in the integral.

Apart from the dx (which I corrected in the quotation in my response), the substitution, ignoring the limits, is right. But as Sutupidmath has correctly followed up, this is a futile approach if one wants to find g'(x) anyway!

 

[Mark44]

Apart from the dx (which I corrected in the quotation in my response), the substitution, ignoring the limits, is right. But as Sutupidmath has correctly followed up, this is a futile approach if one wants to find g'(x) anyway!

My mistake.

 

[NoMoreExams]

If you need to find g'(x) then remember this

[tex] f(x) = \int_{\alpha(x)}^{\beta(x)} \zeta(t) dt \Rightarrow f'(x) = \zeta(\beta(x)) \cdot \beta'(x) - \zeta(\alpha(x)) \cdot \alpha'(x) [/tex]

Hope that helps.

 

[kamranonline]

Thanks alot. I got it now ! :) A very useful rule!