Question

Hello Budiess . I`ve this one chalenging Integral for evaluate but I not solved it. Someboby can help me please ?

integral (from 0 to 2*pi) 18dt/(16+65*(sin t)^2)

Answer 1

this definitely is not a easy problem...even TI-89 giving a weird answer

Answer 2

\[\int\limits_{0}^{2\pi} 18/(16+65 \sin^2(t)) dt\]

Answer 3

The answer is equal \[\Pi\] firend.
...But other way ...dt is on NUMERATOR `n not on denominator

Answer 4

I know....dt there just oout of everything from that fraction...you said the answer is Pi?

Answer 5

yes

Answer 6

From Mathematica Home Edition:\[\text{Timing}\left[\int\limits_0^{2\pi } \frac{18}{16+65 \text{Sin}[t]^2} \, dt \right]=\{0.032884,\pi \} \]Solution time in seconds on a mid-2010 IMac. Pi is the numerical result.

Answer 7

well, before putting [a,b] for it...the intergral should be [tan^-1(9tan(t)/4)]/2 fromt Wolfram Calculator
But i'm finding the way how they get that

Answer 8

Look: http://integrals.wolfram.com/index.jsp?expr=18%2F%2816+%2B+65*sin^2%28x%29%29&random=false

Answer 9

\[\int\limits \frac{18}{16+65 \text{Sin}[t]^2} \, dt=\frac{1}{2} \text{ArcTan}\left[\frac{9 \text{Tan}[t]}{4}\right] \]Intermediate step are not shown in this edition

Answer 10

@rotobey
the square is on sin `n not on t. like that
\[\sin ^{2}t\]

Answer 11

i think i got it...but sorry gotta go now, my friend came to pick up....if no one post it later, i'll come back on post it on for you my solution...

Answer 12

the hint is....try to divided everything by cos t first...then you will have tan^t appear....substitue u = tan t => du = sec^t dt....
use trignometry in the triangle...u = tan t then sec t = square root ( u^2 + 1) => sec ^2 t = u^2 +1 ....

Answer 13

ok geekGuy....

Answer 14

oops du= sec^2 t dt

Answer 15

@geek
IT`s still not clarify so much for me yet friend....

Answer 16

The substitution x = 2 arctan(z), dx = 2 dz/(1 + z^2) will reduce the integrand to a rational function.

Answer 17

like this, divide everything by cos t, you will have

integral ( 18dt/ cos^2(t) / [16/ cos^2(t) + 65 sin^2(t)/cos^2(t)}
integral ( 18sec^2(t)dt/ [16sec^2t + 65tan^2(t)]

then substitute u= tant => du= sec^2 t dt
then you will have
integral ( 18du/ (16 sec^2 +65 u^2))
now imagine tangent is opp/adj so u= tan t means u is opposite and 1 is adj
if you draw a triangle...you will see your hyponeous = square root ( u^2 +1)
cos t = adj/ hyp = 1/ square root( u^2 +1) or sec t= square root( u^2 +1)
therefore sec^2 t= u^2 +1

then you can substute sec^2 t to 16 sec^2t

integral [18du/(16(u^2 +1) +65u^2)]
integral(18du/(16 + 16u^2 + 65u^2))
integral(18du/(16+81 u^2) )

till this point...pretty much done, turn to the back of your calculus book and look for the formula of this kinda of integral (du/ a^2 + u^2) and apply it here...

Answer 18

so it actually dividing cos^2 (t)...not cos(t)....sorry...typing in arush

Answer 19

but divide all by cos^2(t) U modify integrand ?! One better idea wouldn`t mutiply the expression by \[\cos ^{2} t \div \cos ^{2}t\] ?

Answer 20

So seem like you got the way to solve for the answer?

Answer 21

probably you already got it, but i just want to finish last part
Substitute again, v = 9u => dv=9du and v^2=81u^2

\[\int\limits_{0}^{2\pi} (18dv/9)/(4^{2} +v^{2})\]
\[18\tan-^{1} (v/4)/ (9*4)\]
\[\tan-^{1}(9\tan(t)/4)/2\] 0 -> 2pi

Answer 22

@Pavl.BRA
" the square is on sin `n not on t." \[\left\{\text{Sin}[x]^2,\text{Sin}^2[x]\right\}\text{/.}x\to \frac{\pi }{6} \] The expression inside of the first set of braces is a sequence of the symbols, Cap s,i,n,left bracket,x,right bracket,super script 2,comma, Cap s,super script 2,n,left bracket,x and right bracket. When the characters are parsed, compiled and evaluated after x is replaced by pi/6, 30 degrees, the result is posted as the following:\[\left\{\frac{1}{4},\text{Sin}^2\left[\frac{\pi }{6}\right]\right\} \]Starting from the left, 1/4 looks correct, sin 30 degrees is 1/2, 1/2 squared is 1/4. However, the second result is equal to the input form, except for x, which has been replaced by pi/6. This result says that Mathematica has gone as far as it can based on the internal evaluation rules programmed in it. In other words the Wolfram Research management/software developers have elected not to implement the form of the Sine of x in question even though it is the expected form in use and taught everyday.

The preceding comments relate only to the Mathematica application program. In the case of WolframAlpha.com they are attempting to parse and make sense of any sequence of characters thrown at it.

Answer 23

This integral can be solved with pencil and paper. Multiply numerator and denominator by (sec(t))^2, then express the denominator in terms of (tan(t))^2. Substitute u = tan(t), du = (sec(t)^2) dt to reduce the integral to a recognizable form. If all goes well, you will get pi as the answer.