Precalculus Solutions | Expert Assistance for Problems & Examples

Evaluating $lim_{x \to 0} \frac{x - \tan^{- 1} x}{x \sin x}$

Stoockiltj5 2022-01-23

When do we have to change the limits?
$\int_{0}^{1} \sqrt{x - x^{2}} d x$
completing the square gives me:
$\int \sqrt{- x^{2} + x - \frac{1}{4} + \frac{1}{4}} d x$
$\int \sqrt{- (x - \frac{1}{2})^{2} + \frac{1}{4}} d x$
$\int \sqrt{- u^{2} + \frac{1}{4}} d u$
$\int \sqrt{\frac{1}{4} - u^{2}} d u$
Is trig sub the best way to go?
if $u = \frac{1}{2} \sin θ$ then $d u = \frac{1}{2} \cos θ d θ$
$= \int \sqrt{\frac{1}{4} - \frac{1}{4} \sin^{2} θ} \frac{1}{2} c o s θ d θ$
$\int \frac{1}{2} c o s θ \frac{1}{2} c o s θ d θ$
$\frac{1}{4} \int c o s^{2} θ d θ$
using half angle identity I eventually get:
$\frac{1}{4} \int \frac{1}{2} (1 + c o s (2 θ)) d θ$
$\frac{1}{8} \int (1 + c o s (2 θ) d θ$
$\frac{1}{8} (θ + \frac{1}{2} s i n 2 θ)$
Where do I go from here?

Octavio Miller 2022-01-23

Is ${tanh}^{- 1} (\sin 2 (x + d \frac{π}{4})) = \frac{1}{π} \ln (\cot^{2} x)$ ?
As you see the above graph that $\arctan h (\sin 2 (x + \frac{π}{4}))$ is similar to $\frac{1}{π} \ln (\cot^{2} x)$
Does this mean that by some minor alterations in the functions, they would be equal?
Why are both of functions so similar?

Franco Franco 2022-01-23

If $m^{2} \cos \frac{2 π}{15} \cos {\frac{4 π}{15}} \cos {\frac{8 π}{15}} \cos {\frac{14 π}{15}} = n^{2}$ then find the value of $\frac{m^{2} - n^{2}}{n^{2}}$

Keaton Good 2022-01-23

Proving $\int_{0}^{\infty} x^{n} e^{- t x} \frac{\sin x}{x} d x = \frac{\sin n θ}{{(1 + t^{2})}^{\frac{n}{2}}} (n - 1)!$ where $θ = \arcsin \frac{1}{\sqrt{1 + t^{2}}}$
Now I didnt

William Montgomery 2022-01-23

If $x = a {\cos t}^{3}, y = b {\sin t}^{3}$ then what is $\frac{d^{3} y}{{d x}^{3}}$ ?

Anika Osborne 2022-01-23

Let $θ \in (0, \frac{π}{4})$ and $t_{1} = (\tan θ)^{\tan θ}, t_{2} = (\tan θ)^{\cot θ}, t_{3} = (\cot θ)^{\tan θ}$ and $t_{4} = (\cot θ)^{\cot θ}$ , then show that $t_{4} > t_{3} > t_{1} > t_{2}$ .

iocasq4 2022-01-23

Find the smallest value of $f (x) = (\frac{19}{+} \frac{32}{\sin (x)}) (\frac{1}{32} + \frac{9}{\cos x})$ on the interval $(0, \frac{π}{2})$

Anika Klein 2022-01-23

If $\cos (n π) = {(- 1)}^{n}$ , then why is $- \frac{4}{n} \cos (n π) = \frac{4}{n} {(- 1)}^{n + 1}$ ?
I lectures has this written down in one of the solutions to an exercise however I'm not sure how he got $\frac{4}{n} {(- 1)}^{n + 1}$ ? I thought it would just be $- \frac{4}{n}$ . ${(- 1)}^{n}$ ? Am I missing something?

Alisha Pitts 2022-01-23

How do I prove $\tan 1 < \frac{π}{2}$ ?
Prove that the equation
$\sin x \sin (\sin x) = \frac{π}{2} \cos (\sin x)$
Let $t = \sin x, - 1 \leq t \leq 1$ . Then the expression above is equvalent to $t \sin t = \frac{π}{2} \cos t$ . As the function $f (t) = t \sin t - \frac{π}{2} \cos t$ is even, and t=0 is not a solution, I have to prove that f(t) has no positive roots (t>0). So, for the left side $0 < t \leq 1$ and $0 < \sin t \leq \sin 1$ , then $t \sin t \leq \sin 1$ . For the right side $\cos t \geq \cos 1$ , so $\frac{π}{2} \cos t \geq \frac{π}{2} \cos 1$ . The objective is to prove that $\sin 1 < \frac{π}{2} \cos 1$ , or, equivalently, $\tan 1 < \frac{π}{2}$

Jacquelyn Sanders 2022-01-23

If $z = \sqrt{3} + i = (a + i b) (c + i d)$ , then find $\tan^{- 1} \frac{b}{a} + \tan^{- 1} \frac{d}{c}$

Rudy Koch 2022-01-23

Solving $\sin 2 x \sin x + \cos^{2} x = \sin 5 x \sin 4 x + \cos^{2} 4 x$

estadsocm3 2022-01-23

Trigonometric Equation: $4 \sin θ = 3 \tan θ$
Solve this equation for $- 180^{\circ} \leq θ \leq 180^{\circ}$

Wendy Gutierrez 2022-01-23

How does $\frac{- e^{- x}}{\sqrt{e^{- 2 x} - 1}}$ rearrange?

Turnseeuw 2022-01-23

$\sin (\arccos (\frac{1}{\sqrt{3}})) = \frac{\sqrt{2}}{\sqrt{3}}$ proof this equation

Kendall Holder 2022-01-23

$\sin (x) + p \cos (x) = 2 p, p \in R$ has solutions for p=?

Jamarion Kerr 2022-01-23

For proving
$\frac{16}{\cos 4 x + 7} = \frac{1}{\sin^{4} x + \cos^{2} x} + \frac{1}{\sin^{2} x + \cos^{4} x}$
I tried to use that:
$\sin^{4} x + \cos^{4} x = \sin^{4} x + 2 \sin^{2} x \cos^{2} x + \cos^{4} x - 2 \sin^{2} x \cos^{2} x$
$= {(\sin^{2} x + \cos^{2} x)}^{2} - 2 \sin^{2} x \cos^{2} x$
$= 1^{2} - {\frac{12}{2 \sin x \cos x}}^{2}$
$= 1 - {\frac{12}{\sin}}^{2} (2 x)$
$= 1 - \frac{12}{\frac{1 - \cos 4 x}{2}}$
$= \frac{34}{+} \frac{14}{\cos 4} x$

pozicijombx 2022-01-23

prove that $2 \arctan (\csc \arctan x - \tan a r o t x) = \arctan x$

Jenny Branch 2022-01-23

How to prove : $\cos^{3} 2 θ + 3 \cos 2 θ = 4 (\cos^{6} θ - \sin^{6} θ)$

Kristina Mcclain 2022-01-22

How do you write the composite function in the form $f (g (x)) y = \sin (7 x)$

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