Use the substitution x = 2 − cos θ to evaluate the integral ∫ 2 3/2 ( x − 1 3 − x )1 2 dx. Show that, for a < b, ∫ q p ( x − a b − x )1 2 dx = (b − a)(π + 3√ 3 − 6) 12 , where p =

If the integral as written in my comment is accurate, then we have

Expand the polynomial, then complete the square within the square root:

Let and :

Recall that for all , but for all in the integration interval we have . So :

Recall the double angle identity,

You can determine the more general result in the same way.

Complete the square to get

and let for brevity. Note that

Make the following substitution,

and the integral reduces like before to

where

(Depending on the interval [p, q] and thus [P, Q], the square root of cosine squared may not always reduce to sine.)

Resolving the integral and replacing c, with

because , gives

Without knowing p and q explicitly, there's not much more to say.

∠1≅∠2 by the alternate exterior angles theorem.

step-by-step explanation:

given, a ∥ b and ∠1 ≅ ∠3 .we have to prove that e ∥ f

we know that ∠1≅∠3 and that a || b because they are given. we see that by the alternate exterior angles theorem. therefore, ∠2≅∠3 by the transitive property. so, we can conclude that e || f by the converse alternate exterior angles theorem.

we have to fill the missing statement.

transitivity property states that if a = b and b = c, then a = c.

now, given ∠1≅∠3 and by transitivity property ∠2≅∠3 .

hence, to apply transitivity property one angle must be common which is not in result after applying this property which is ∠1.

the only options in which ∠1 is present are ∠1 and ∠2, ∠1 and ∠4

∠1 and ∠4 is not possible ∵ after applying transitivity we didn't get ∠4.

hence, the missing statement is ∠1≅∠2.

so, ∠1≅∠2 by the alternate exterior angles theorem.

4.

step-by-step explanation:

it have the same area.brainliest