Can someone explain to me how they got 1/2?

I'm having trouble with #2. I've got it down to the part where it would be the integral of 5cos^3(pheta)/sin(pheta). I'm not sur

Butoxors [25]

$\displaystyle\int\frac{\sqrt{25-x^2}}x\,\mathrm dx$

Setting $x=5\sin\theta$ , you have $\mathrm dx=5\cos\theta\,\mathrm d\theta$ . Then the integral becomes

$\displaystyle\int\frac{\sqrt{25-(5\sin\theta)^2}}{5\sin\theta}5\cos\theta\,\mathrm d\theta$
$\displaystyle\int\sqrt{25-25\sin^2\theta}\dfrac{\cos\theta}{\sin\theta}\,\mathrm d\theta$
$\displaystyle5\int\sqrt{1-\sin^2\theta}\dfrac{\cos\theta}{\sin\theta}\,\mathrm d\theta$
$\displaystyle5\int\sqrt{\cos^2\theta}\dfrac{\cos\theta}{\sin\theta}\,\mathrm d\theta$

Now, $\sqrt{x^2}=|x|$ in general. But since we want our substitution $x=5\sin\theta$ to be invertible, we are tacitly assuming that we're working over a restricted domain. In particular, this means $\theta=\sin^{-1}\dfrac x5$ , which implies that $\left|\dfrac x5\right|\le1$ , or equivalently that $|\theta|\le\dfrac\pi2$ . Over this domain, $\cos\theta\ge0$ , so $\sqrt{\cos^2\theta}=|\cos\theta|=\cos\theta$ .

Long story short, this allows us to go from

$\displaystyle5\int\sqrt{\cos^2\theta}\dfrac{\cos\theta}{\sin\theta}\,\mathrm d\theta$

to

$\displaystyle5\int\cos\theta\dfrac{\cos\theta}{\sin\theta}\,\mathrm d\theta$
$\displaystyle5\int\dfrac{\cos^2\theta}{\sin\theta}\,\mathrm d\theta$

Computing the remaining integral isn't difficult. Expand the numerator with the Pythagorean identity to get

$\dfrac{\cos^2\theta}{\sin\theta}=\dfrac{1-\sin^2\theta}{\sin\theta}=\csc\theta-\sin\theta$

Then integrate term-by-term to get

$\displaystyle5\left(\int\csc\theta\,\mathrm d\theta-\int\sin\theta\,\mathrm d\theta\right)$
$=-5\ln|\csc\theta+\cot\theta|+\cos\theta+C$

Now undo the substitution to get the antiderivative back in terms of $x$ .

$=-5\ln\left|\csc\left(\sin^{-1}\dfrac x5\right)+\cot\left(\sin^{-1}\dfrac x5\right)\right|+\cos\left(\sin^{-1}\dfrac x5\right)+C$

and using basic trigonometric properties (e.g. Pythagorean theorem) this reduces to

$=-5\ln\left|\dfrac{5+\sqrt{25-x^2}}x\right|+\sqrt{25-x^2}+C$

4 0

2 years ago

Read 2 more answers

10(a-2.5+0.56b) is this equivalent to 10a - 25+56b

kirill115 [55]

Hey there!

10(a - 2.5 + 0.56b)

= 10(a) + 10(-2.5) + 10(0.56b)

= 10a - 25 + 5.6b

= 10a + 5.6b - 25

Therefore, your answer is: 10a + 5.6b - 25

Good luck on your assignment and enjoy your day!

~Amphitrite1040:)

3 0

2 years ago

Pizza take 20 minutes to cook. For pizzas can be cooked at one time in the pizza oven. How long would it take to cook 10 pizzas

ZanzabumX [31]

You'd fine the unit rate, in the context of the problem it would be how many minutes it takes for 1 pizza to cook
20/4= 5 minutes for 1 pizza
we are finding how long it would take to cook 10
5 times 10= 50, so 50 minutes

8 0

2 years ago

I seriously need help. Please help. Be brainliest answer

WITCHER [35]

12a) yes, there will be two vanilla cupcakes left over after the baker makes ten more.

7 0

2 years ago

Which property do i use to solve x-10=15

Stells [14]

Add both sides by 10

x=25

3 0

3 years ago

Can someone explain to me how they got 1/2?​

Can someone explain to me how they got 1/2?