1. -1 points LarCalc10 5.6.009 My Notes Ask Your Evaluate the expression without

Question

1. -1 points LarCalc10 5.6.009 My Notes Ask Your Evaluate the expression without using a calculator arccot( v3) 2. -2 points LarCalc10 56.022 My Notes Ask Your Evaluate each expression without using a calculator. [Hint: Make a sketch of a right triangle.] (a) tan(arccos(f)) N2 (b) cm(arcsin( 3. -41 points LarCalc10 5.6.032 My Notes Ask Your Write the expression in algebraic form. [Hint: Sketch a right triangle.] cs (arctan( ) poits LarCalc10 5.7006 My Notes Ask Your Find the indefinite integral. (Use C for the constant of integration.) x 16 11 points LarCalc10 5.7013 My Notes Ask Your Find the indefinite integral. (Use C for the constant of integration.) V16-tan2x

Explanation / Answer

arccot(-sqrt3)

This is same as arctan(-1/sqrt3) where we just reciprocal-ed it up.

We know that arctan(1/sqrt3) is pi/6

So, using arctan(-x) = -arctan(x),
we can tell that arctan(-1/sqrt3) = -pi/6

So -pi/6

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2)
arccos(sqrt2/2) is pi/4
tan(pi/4) = 1

1

b)
cos(arcsin(5/13))

With arcsin(5/13), draw a right triangle with opp = 5 and hyp = 13
cuz sin = opp/hyp from soh in sohcahtoa...

With that, adj = sqrt(13^2 - 5^2) ---> adj = 12

And we know that cos(...) = adj/hyp
which then becomes
12/13

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3)
With arctan(x/sqrt2), we have
opp = x and adj = sqrt2
So, hypotenuse = sqrt(x^2 + 2)

So, csc(..) is hyp/opp

So, it becomes

sqrt(x^2 + 2) / x

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4)
Take u = x - 7
So, deriving, du/dx = 1
And thus, du = dx

Subbing it in :
(integral) 1/(16 + u^2) * du

This is of the form integral of 1/(a^2 + x^2) which is
1/a * arctan(x/a)

So, here we have
1/4 * arctan(u/4) + C

Then sub back u = x - 7 :

1/4 * arctan((x-7)/4) + C ----> ANSWER

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5)
u = tanx
du = sec^2xdx

So, integral becomes :

integral du / (sqrt(16 - u^2)
which is of the form dx/sqrt(a^2 - x^2), which is arcsin(x/a)

So, here, we have :
arcsin(u/4)+ C

arcsin(tan(x)/4) + C ----> ANSWER

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