Integration of trigonometry (1 Viewer)

[atar90plus]

Hello Could you guys please help me with this question. Please show full working and solution. Please note that the answer (labelled gif) is 64/3 and the answer (labelled codecogs) is 1/3

Thanks

 

[Drongoski]

cannot read the question.

 

[Shadowdude]

Uhh, why can't you use the substitution u = 16-x^2

 

[Shadowdude]

btw, the question is:




And then at the end it's like: "where " for some reason

 

[atar90plus]

Yeah sorry this is my first time using latex. The part that say "where" is the condition given. The question tells us what to let x equal to. Sory I am only in year 11 that is why i still need help for this topic.

 

[atar90plus]

Yes shadowdude is right. This is the question

btw, the question is:




And then at the end it's like: "where " for some reason

 

[Drongoski]

Thanks Shadowdude.

Then the integral is simply =



Are you using x = 4sin @ as the substitution?

 

[Shadowdude]

In future, don't link images of tex. Use 'tex' tags. [tex.] and [./tex]

remove the full stops and paste your code between them

 

[atar90plus]

Could you please show me step by step how to get your solution because I need to know how to do this question in order to solve the other ones.
Thanks

Thanks Shadowdude.

Then the integral is simply =

 

[atar90plus]

Yes. I really need your help with this question if you dont mind

Thanks Shadowdude.

Then the integral is simply =



Are you using x = 4sin @ as the substitution?

 

[atar90plus]

How did you get the part where it says -1/3 and the part of the root cubed. I did not get that in my working out. I believe I did it wrong so that is why I need the step by step solution so I can solve the other questions and see where I went wrong.

 

[RishBonjour]

I got the same answer as Drognoski,
I just did it like a normal 3u integration. dx/dtheta, and then when x=0, theta=... etc. and then you sub your stuf in

my second last line was
64 times the integral of sin(theta)cos^(theta) d(theta) (from 0, to pi/2)

you then integrate the above, and should get you the answer.

 

[Timske]

​<a href="http://www.codecogs.com/eqnedit.php?latex=\int_{0}^{4} x\sqrt{16 -x^2} dx ~~,~~ x = 4sin\theta \\\\ \int_{0}^{\frac{\pi}{2}} 16sin\theta cos\theta\sqrt{16-16sin^2\theta}~d\theta \\\\ = 64\int_{0}^{\frac{\pi}{2}} sin\theta cos\theta\sqrt{1 - sin^2\theta} ~d\theta \\\\ = 64\int_{0}^{\frac{\pi}{2}} sin\theta cos\theta\sqrt{cos^2\theta} ~d\theta \\\\ = 64\int_{0}^{\frac{\pi}{2}} sin\theta cos^2\theta ~d\theta ~,~ sin\theta~cos^2\theta\Rightarrow \left (\frac{sin3\theta @plus; sin\theta}{4} \right )" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\int_{0}^{4} x\sqrt{16 -x^2} dx ~~,~~ x = 4sin\theta \\\\ \int_{0}^{\frac{\pi}{2}} 16sin\theta cos\theta\sqrt{16-16sin^2\theta}~d\theta \\\\ = 64\int_{0}^{\frac{\pi}{2}} sin\theta cos\theta\sqrt{1 - sin^2\theta} ~d\theta \\\\ = 64\int_{0}^{\frac{\pi}{2}} sin\theta cos\theta\sqrt{cos^2\theta} ~d\theta \\\\ = 64\int_{0}^{\frac{\pi}{2}} sin\theta cos^2\theta ~d\theta ~,~ sin\theta~cos^2\theta\Rightarrow \left (\frac{sin3\theta + sin\theta}{4} \right )" title="\int_{0}^{4} x\sqrt{16 -x^2} dx ~~,~~ x = 4sin\theta \\\\ \int_{0}^{\frac{\pi}{2}} 16sin\theta cos\theta\sqrt{16-16sin^2\theta}~d\theta \\\\ = 64\int_{0}^{\frac{\pi}{2}} sin\theta cos\theta\sqrt{1 - sin^2\theta} ~d\theta \\\\ = 64\int_{0}^{\frac{\pi}{2}} sin\theta cos\theta\sqrt{cos^2\theta} ~d\theta \\\\ = 64\int_{0}^{\frac{\pi}{2}} sin\theta cos^2\theta ~d\theta ~,~ sin\theta~cos^2\theta\Rightarrow \left (\frac{sin3\theta + sin\theta}{4} \right )" /></a>

 

[Timske]

<a href="http://www.codecogs.com/eqnedit.php?latex=\int_{0}^{2} \frac{dx}{1 @plus; x^2} ~,~x=2tan\theta \\\\ \frac{1}{4}\int_{\frac{\pi}{4}}^{0} \frac{2sec^2\theta }{1 @plus; tan^2\theta}~d\theta ~,~ 1 @plus;tan^2\theta \Rightarrow sec^2\theta \\\\ = \frac{1}{4}\int_{0}^{\frac{\pi}{4}} 2~d\theta \\\\ = \frac{1}{2}\left [ \theta\right ]_{0}^{\frac{\pi}{4}} \\\\ = \frac{1}{2}(\frac{\pi}{4}) = \frac{\pi}{8} \approx \frac{1}{3}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\int_{0}^{2} \frac{dx}{1 + x^2} ~,~x=2tan\theta \\\\ \frac{1}{4}\int_{\frac{\pi}{4}}^{0} \frac{2sec^2\theta }{1 + tan^2\theta}~d\theta ~,~ 1 +tan^2\theta \Rightarrow sec^2\theta \\\\ = \frac{1}{4}\int_{0}^{\frac{\pi}{4}} 2~d\theta \\\\ = \frac{1}{2}\left [ \theta\right ]_{0}^{\frac{\pi}{4}} \\\\ = \frac{1}{2}(\frac{\pi}{4}) = \frac{\pi}{8} \approx \frac{1}{3}" title="\int_{0}^{2} \frac{dx}{1 + x^2} ~,~x=2tan\theta \\\\ \frac{1}{4}\int_{\frac{\pi}{4}}^{0} \frac{2sec^2\theta }{1 + tan^2\theta}~d\theta ~,~ 1 +tan^2\theta \Rightarrow sec^2\theta \\\\ = \frac{1}{4}\int_{0}^{\frac{\pi}{4}} 2~d\theta \\\\ = \frac{1}{2}\left [ \theta\right ]_{0}^{\frac{\pi}{4}} \\\\ = \frac{1}{2}(\frac{\pi}{4}) = \frac{\pi}{8} \approx \frac{1}{3}" /></a>

 

[Drongoski]