Mathematics: Evaluate dx / ^ 9-8x - x2^ a. ln (-8 - 2x) + c b. sin -1 (x+4/5) + c c. sin -1 (x-4/5) + c d. ln ^ 9-8x-x2^ + c my apologies if i wrote it out a little wrong and you can’t understand! hopefully i did it right

D)30 is the answer sorry if I get it wrong but I think it’s the answer...

Evaluate dx / ^ 9-8x - x2^ a. ln (-8 - 2x) + c b.

Ответ:

gomek

14.03.2022

It depends on what you mean by the delimiting carats "^"...

Since you use parentheses appropriately in the answer choices, I'm going to go out on a limb here and assume something like "^x^" stands for $\sqrt x$ .

In that case, you want to find the antiderivative,

$\displaystyle\int\frac{\mathrm dx}{\sqrt{9-8x-x^2}}$

Complete the square in the denominator:

9-8x-x^2=25-(16+8x+x^2)=5^2-(x+4)^2

Now substitute $x+4=5\sin y$ , so that $\mathrm dx=5\cos y\,\mathrm dy$ . Then

$\displaystyle\int\frac{\mathrm dx}{\sqrt{9-8x-x^2}}=\int\frac{5\cos y}{\sqrt{5^2-(5\sin y)^2}}\,\mathrm dy$

which simplifies to

$\displaystyle\int\frac{5\cos 
y}{5\sqrt{1-\sin^2y}}\,\mathrm dy=\int\frac{\cos y}{\sqrt{\cos^2y}}\,\mathrm dy$

Now, recall that $\sqrt{x^2}=|x|$ . But we want the substitution we made to be reversible, so that

$x+4=5\sin y\iff y=\sin^{-1}\left(\dfrac{x+4}5\right)$

which implies that $-\dfrac\pi2\le y\le\dfrac\pi2$ . (This is the range of the inverse sine function.)

Under these conditions, we have $\cos y\ge0$ , which lets us reduce $\sqrt{\cos^2y}=|\cos y|=\cos y$ . Finally,

$\displaystyle\int\frac{\cos y}{\cos y}\,\mathrm dy=\int\mathrm dy=y+C$

and back-substituting to get this in terms of

yields

$\displaystyle\int\frac{\mathrm dx}{\sqrt{9-8x-x^2}}=\sin^{-1}\left(\frac{x+4}5\right)+C$

0,0(0 оценок)

Evaluate dx / ^ 9-8x - x2^ a. ln (-8 - 2x) + c b. sin -1 (x+4/5) + c c. sin -1 (x-4/5) + c d. ln ^ 9-8x-x2^ + c my apologies if i wrote it out a little wrong and you can’t understand! hopefully i did it right

Ответ:

Ответ:

Ask your question to the AI advisor

More tips

Answers on questions: Mathematics

Ask an AI advisor a question