Question

Given that \(\large \color{green}{\int\limits_0^4(x^3 \cdot \sqrt{9 + x^2}).dx = a}\), what is the value of \(\large\color{blue}{[a]}\), where [] is Greatest Integer Function...

Answer 1

*

Answer 2

\[\large \color{green}{\int\limits\limits_0^4(x^3 \cdot \sqrt{9 + x^2}).dx }\]
let \[\large x = 3 \tan \theta\] \[\large dx = 3 \sec^2 \theta d \theta\]
\[\large \int\limits\limits\limits\limits\limits_0^4(3 \tan \theta)^3 \cdot \sqrt{9 + (3 \tan \theta )^2} (3 \sec^2 \theta d \theta)\]

Answer 3

Sorry, then??

Answer 4

Take two..

Answer 5

\[\large \int\limits\limits\limits\limits\limits\limits_0^4(3 \tan \theta)^3 \cdot \sqrt{9 + (3 \tan \theta )^2} (3 \sec^2 \theta d \theta) = \]
\[\large 27 \times 3 \times \int\limits_{0}^{4} \tan^3 \theta \cdot \sqrt{9(1+ \tan^2 \theta )} \sec^2 \theta d \theta)\]

follow to here?

Answer 6

\[\large 81 \int\limits\limits_{0}^{4} \tan^3 \theta \cdot \sqrt{9\sec^2 \theta } \sec^2 \theta d \theta) =\] \[\large 243 \int\limits\limits\limits_{0}^{4} \tan^3 \theta \cdot \sec \theta \sec^2 \theta d \theta) =\]
\[\large 243 \int\limits\limits\limits\limits_{0}^{4} \tan^3 \theta \cdot \sec^3 \theta d \theta =\]

Answer 7

I never get tired of this...
Miss Singh is missing again :3

Answer 8

Oh crap i kept the limits 0 to 4, ah well just ignore those till the end I guess.

Answer 9

Now use tan^2 = sec^2 - 1\[\large 243 \int\limits\limits\tan^3 \theta \cdot \sec^3 \theta d \theta = 243 \int\limits\limits (\sec^2 \theta -1) \tan \theta \sec^3 \theta d \theta =\]
\[\large 243 \int\limits\limits\limits (\sec^2 \theta -1) \tan \theta \sec^3 \theta d \theta = 243 \int\limits \sec^5 \theta \tan \theta - \tan \theta \sec^3 \theta d \theta \] Make yet another substitution (wondering if there was a faster way to do this... by parts?)

\[u = \sec \theta \]
\[du = \tan \theta \sec \theta d \theta \]

Answer 10

No need o.O
\[\large 243\int\limits_{0}^{4}\tan^2\theta\sec^2\theta(\sec\theta\tan\theta)d\theta\]\[\large 243\int\limits_{0}^{4}(\sec^2\theta-1)\sec^2\theta(\sec\theta\tan\theta)d\theta\]

Answer 11

And you'll get \[\large \left[ \frac{\sec^5 \theta}{5} - \frac{ \sec^3 \theta} {3}\right]\]

Answer 12

Not yet done.
\[\large x=3\tan\theta\]\[\huge \theta=\frac13\tan^{-1}x\]

Answer 13

I know :P x=3tanθ and x=0 to x=4

Answer 14

Yeah... but integrating was the easy part.
Now, we still have to evaluate it :D

Answer 15

If it was an indefinite integral, you can change it back to non-theta-crap with a right triangle

\[\cos \theta = \frac {3} {\sqrt{9+x^2}}\] so \[\large \frac{ 1 }{ \cos \theta } = \sec \theta = \frac {\sqrt{9+x^2}}{3}\] and then put that back into...

\[\large \left[ \frac{\sec^5 \theta}{5} - \frac{ \sec^3 \theta} {3}\right] = \]

\[\LARGE \left[ \frac{(\sqrt{9+x^2})^5 }{3^5 \times 5} - \frac{ (\sqrt{9+x^2})^3 } {3^3 \times 3}\right]_{0}^{4}\]

Answer 16

Oh, throw a 243 in front of that, i think we dropped it.
\[\LARGE 243 \left[ \frac{(\sqrt{9+x^2})^5 }{3^5 \times 5} - \frac{ (\sqrt{9+x^2})^3 } {3^3 \times 3}\right]_{0}^{4}\]

Answer 17

Which, thank god, agrees with the number wolfram alpha gets!

Then you just use the greatest integer function, so you essentially just round down.

Answer 18

Miss Singh ??? @terenzreignz

Answer 19

Your profile says "Gurpreet Singh" after all. Also, puns ^.^

Answer 20

What is the idea behind 'Miss' there??

Answer 21

I don't know, you said you were a girl, over chat?
Besides, Miss Singh sound so much awesomer and punnier :3

Answer 22

*sounds