Veeky Forums can't find it

trig sub it prolly

can't you just regular sub it?

Depends what you mean by "regular". I mean, you could do any substitution but that won't necessarily yield successful integration.

...

This

sin(2016x) = 1-u
integrate sqrt(u)dx for x = 0 to pi/2
(2/3)u^(3/2)
(2/3)(1-sin(1008pi))^(3/2) - (2/3)(1-sin(0))^(3/2)
(2/3)-(2/3)
0

>integrate sqrt(u)dx for x
you mean integrate sqrt(u)dx (du/dx) for u

Is this bait or is Veeky Forums actually this retarded?

>integral of a positive non-null function is 0

right? I hope it's actually bait.

[math]\sqrt{1 - sin(2016x)} = \sqrt{1} - \sqrt{sin(2016x)} = 1 - (1 - \sqrt{cos(2016x)}) = \sqrt{cos(2016x)}[/math]

then you just substitute and solve. This is freshman year math.

Why not 2017x? Why are you posting an year old equation?

Because the trick to solve it only works because 2016 is even.

[eqn] 1 - \sin(2z) = \left( \sin(z) - \cos(z) \right)^2 [/eqn]

kys

>spoonfeeding OP

Multiply above and bellow by [math] \sqrt{1+\sin(2016x)} [/math] .

you should probably do some meditation, it would help with your anger issues.

I know you engineers have trouble with math, but come on.

Mathematica 11 or 10?

Just use a double-angle identity. Very easy.

Easy.
cos^2 x = (1-sinx)(1+sinx) so the integrand is
+/- cosx/sqrt(1+sinx)
and then use the substitution u= 1+sinx.

no it's highschool math

Found it:
[math] \mathscr{E} [/math]

>doesn't even take 30 seconds out of his day to evaluate the derivative of his answer and find it is wrong
>doesn't realize the reason for this is that
[math]\sqrt{1-\sin{x}} = \sqrt{1}-\sqrt{\sin{2016 x}}[/math] is a completely illegitimate operation and completely fucking stupid
>still has the arrogance to call OP a freshman, some post from above bait, and some post calling out his wrong answer a post made by someone with anger issues and braintletism
>is a fucking tripfag on top of all of that

right? I hope it's actually bait.

Fuck you for making me think about that for even a second.

how is it illegitimate to expand a function?

I took this course twice now, at a top university no less, so I think I know what I'm talking about.

Ignore all tripfags.

Hello newfriend

Look at the file name you fucking retard

Then feel free to lecture me with how, for example, you can use your "legitimate" method of expanding radicals to imply the Pythagorean Theorem can be rewritten as

[math]a^{2} + b^{2} = c^{2} =>[/math]
[math]\sqrt{a^{2} + b^{2}} = \sqrt{c^{2}} =>[/math]
[math]\sqrt{a^{2}} + \sqrt{b^{2}} = c =>[/math]
[math]a + b = c[/math]

You are right, I should just have let him and his amazing method by himself

the pythagorean theorem applies in 2D, not in 1D where all vectors are colinear in the first place.

This is getting ridiculous, I'm done giving free classes.

>failing a course somehow gives you more legitimacy
lol?

>Being this autistic

Yes.

...

> [math] \sqrt{1 - sin(2016x)} = \sqrt{1} - \sqrt{sin(2016x)} [/math]
L0Lno fgt pls

>pie over 2

Well, i found it. It is on the left side of the equation.

Just so you know, user, your humor isn't lost on everyone. Good job.

>16+9=25 implies 4+3=5
>[math]sqrt{2}=2[/math]
Magnificent.

You're forgetting the astounding result,
sqrt(100) = sqrt(1) + sqrt(1) + sqrt(1) + sqrt(1)... = 100

why not cancell the root with the big S?

my calc is in drugs now

wouldnt it be easier to use the inverse function and then just calculate the area?