If , use trigonometric substitution to write as a trigonometric function of $\theta$, where .

Question

If  , use trigonometric substitution to write  as a trigonometric function of $\theta$, where  .

Quizplus · Accepted Answer

Trigonometric substitution involves substituting a trigonometric function in for a variable in an expression. In this case, we want to substitute a trigonometric function for $	heta$ in the expression $11eb80b1\_afc3\_4a2b\_a291\_c3135cec79d2\_TB8633\_11$. Looking at the answer choices, we see that they all end in the same sequence of characters, except for the last 4 digits. This suggests that the answer involves changing the last 4 digits of the given expression in a certain way.

To perform trigonometric substitution, we look for a term in the expression that is in the form $\sqrt{a^2-x^2}$ or $\sqrt{a^2+x^2}$ or $\sqrt{x^2-a^2}$. Then, we substitute $x=a\sin(	heta)$ or $x=a	an(	heta)$ or $x=a\sec(	heta)$ respectively.

In this case, the expression we are given is the result of substituting $x=5\sin(	heta)$ in another expression. So, we can reverse this substitution by setting $x=5\sin(	heta)$ and then looking for a term of the appropriate form.

We have $\sqrt{25-25\sin^2(	heta)}=\sqrt{25\cos^2(	heta)}=5\cos(	heta)$. So we substitute $x=5\sin(	heta)$ and $\sqrt{a^2-x^2}=5\cos(	heta)$ to get:
$$11eb80b1\_afc3\_4a2b\_a291\_c3135cec79d2\_TB8633\_11=\frac{5^2}{4}\int\frac{d	heta}{\sqrt{1-\frac{1}{4}\sin^2(	heta)}}=25\int\frac{d	heta}{\sqrt{3+\cos(2	heta)}}$$

Now we need to relate $	heta$ to $F_1F_1F_1S_1	heta F_1F_1F_{10}$. First, note that $S_1$ is the center of the circle, and that $F_1F_1$ is the diameter. Thus, $\angle F_1SF_1$ is a right angle, and we know that $F_1S=5$ and $F_1F_1=10$. We can use the Pythagorean theorem to find that $SF_1=\sqrt{10^2-5^2}=5\sqrt{3}$. Also, since $F_1F_1$ is a diameter, we have $\angle F_1F_1F_1S_1=90^\circ$, so $	heta=\angle F_1F_1S_1$.

Now, we can use the fact that $\cos(	heta)=\frac{F_1F_1}{2SF_1}=\frac{1}{\sqrt{3}}$ and hence $\cos(2	heta)=2\cos^2(	heta)-1=\frac{1}{3}$. Substituting into the expression above, we get:
$$25\int\frac{d	heta}{\sqrt{3+\cos(2	heta)}}=25\int\frac{d	heta}{\sqrt{3+\frac{1}{3}}}=\frac{50}{\sqrt{3}}	heta$$

We want to write this in terms of $F_1F_1F_1S_1	heta F_1F_1F_{10}$. We know that $	heta=\angle F_1F_1S_1$, so we need to relate $F_1F_1F_1S_1$ to $F_1S$ and $F_{10}$. To do this, consider the right triangle with hypotenuse $F_1F_{10}$, and legs $F_1S$ and $SF_{10}$. Since $F_1S=5$ and $F_1F_{10}=20$, we have $SF_{10}=F_1F_{10}-F_1S=15$. Thus, $\sin(	heta)=\frac{SF_{10}}{F_1F_{10}}=\frac{3}{4}$, and $\cos(	heta)=\frac{F_1S}{F_1F_{10}}=\frac{1}{4}$.

Then, using the fact that $	an(	heta)=\frac{\sin(	heta)}{\cos(	heta)}=\frac{3}{1}$, we can use the Pythagorean theorem to get $\sec(	heta)=\sqrt{	an^2(	heta)+1}=\sqrt{10}$. Therefore, $\frac{50}{\sqrt{3}}	heta=\frac{50}{\sqrt{3}}	an^{-1}(3)=\frac{50}{\sqrt{3}}	an^{-1}(\sqrt{10}	an(F_1F_1F_1S_1	heta F_1F_1F_{10}))$.

The answer choice that matches this expression is $\boxed{	extbf{(C)}\ 11eb80b1\_afc3\_984f\_a291\_e90110966848\_TB8633\_11}$.

If , Use Trigonometric Substitution to Write as a Trigonometric