Using the substitution $ t = \tan \frac{x}{2} $, otherwise, evaluate $$\int_{\frac{\pi}{3}}^{\frac{\pi}{2}} \frac{1}{3\sin x - 4\cos x + 5} \, dx.$$ The base of a solid is the region bounded by $ y = x^2 $, $ y = -x^2 $ and $ x = 2 $. Each cross-section perpendicular to the $ x $-axis is a trapezium, as shown in the diagram. The trapezium has three equal sides and its base is twice the length of any one of the equal sides. Find the volume of the solid. The point $ S(ae, 0) $ is the focus of the hyperbola $ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 $ on the positive $ x $-axis. The points $ P(at, br) $ and $ Q \left( \frac{a}{t}, -\frac{b}{t} \right) $ lie on the asymptotes of the hyperbola, where $ t > 0 $. The point $ M \left( \frac{a(t^2 + 1)}{2t}, \frac{b(t^2 - 1)}{2t} \right) $ is the midpoint of $ PQ $. (i) Show that $ M $ lies on the hyperbola. (ii) Prove that the line through $ P $ and $ Q $ is a tangent to the hyperbola at $ M $. (iii) Show that $ OP \times OQ = \sqrt{a^2t^2 + b^2t^2} \times \sqrt{\frac{a^2}{t^2} + \frac{b^2}{t^2}}$. (iv) If $ P $ and $ S $ have the same $ x $-coordinate, show that $ MS $ is parallel to one of the asymptotes of the hyperbola.

SSCE HSC Mathematics Extension 2 Integration by substitution: Using the substitution $ t = \ta

Using the substitution $ t = \tan \frac{x}{2} $, otherwise, evaluate $$\int_{\frac{\pi}{3}}^{\frac{\pi}{2}} \frac{1}{3\sin x - 4\cos x + 5} \, dx.$$ The base of a solid is the region bounded by $ y = x^2 $, $ y = -x^2 $ and $ x = 2 $ - HSC - SSCE Mathematics Extension 2 - Question 13 - 2014 - Paper 1

Question 13

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Using the substitution $ t = \tan \frac{x}{2} $, otherwise, evaluate $$\int_{\frac{\pi}{3}}^{\frac{\pi}{2}} \frac{1}{3\sin x - 4\cos x + 5} \, dx.$$ The base of a... show full transcript

Worked Solution & Example Answer:Using the substitution $ t = \tan \frac{x}{2} $, otherwise, evaluate $$\int_{\frac{\pi}{3}}^{\frac{\pi}{2}} \frac{1}{3\sin x - 4\cos x + 5} \, dx.$$ The base of a solid is the region bounded by $ y = x^2 $, $ y = -x^2 $ and $ x = 2 $ - HSC - SSCE Mathematics Extension 2 - Question 13 - 2014 - Paper 1

Step 1

Using the substitution $ t = \tan \frac{x}{2} $, otherwise, evaluate

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Answer

To solve the integral $\int_{\frac{\pi}{3}}^{\frac{\pi}{2}} \frac{1}{3\sin x - 4\cos x + 5} \, dx$ using the substitution $t = \tan \frac{x}{2}$ , we use the formulas:

$\sin x = \frac{2t}{1+t^2}$ and $\cos x = \frac{1-t^2}{1+t^2}$ .

Substituting these into the integral gives: $\int \frac{1}{3\left(\frac{2t}{1+t^2}\right) - 4\left(\frac{1-t^2}{1+t^2}\right) + 5} rac{2}{1+t^2} \, dt$

This expression can be simplified and then evaluated over the relevant limits.

Step 2

Find the volume of the solid.

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Answer

To find the volume of the solid, we first need to establish the area of the cross-section. The trapezium's top and bottom bases can be determined from the intersection points of the curves $y = x^2$ and $y = -x^2$ . The area $A$ of the trapezium can be expressed as: $A = \frac{1}{2} (\text{Base}_1 + \text{Base}_2) \cdot h$

where $h$ is the distance between the bases. Finally, to get the volume $V$ , we integrate this area from $x = 0$ to $x = 2$ : $V = \int_{0}^{2} A \, dx$ .

Step 3

Show that $ M $ lies on the hyperbola.

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To prove that $M \left( \frac{a(t^2 + 1)}{2t}, \frac{b(t^2 - 1)}{2t} \right)$ lies on the hyperbola, substitute the coordinates of $M$ into the hyperbola equation: $\frac{\left(\frac{a(t^2 + 1)}{2t}\right)^2}{a^2} - \frac{\left(\frac{b(t^2 - 1)}{2t}\right)^2}{b^2} = 1$

Simplifying this should lead to confirming its validity.

Step 4

Prove that the line through $ P $ and $ Q $ is a tangent to the hyperbola at $ M $.

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To prove that the line through $P$ and $Q$ is a tangent to the hyperbola at $M$ , we first find the slopes at points $P$ and $Q$ using derivatives. Then, we need to check if the slope calculated at $M$ matches with the slope of the line segment $PQ$ .

Step 5

Show that $ OP \times OQ = \sqrt{a^2t^2 + b^2t^2} \times \sqrt{\frac{a^2}{t^2} + \frac{b^2}{t^2}}$.

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To show this relationship, we derive both lengths $OP$ and $OQ$ from the coordinates of points $P$ and $Q$ . Using the distance formula: $OP = \sqrt{(at)^2 + (br)^2}$

Then we can compute $OP \times OQ$ using the expressions derived earlier.

Step 6

If $ P $ and $ S $ have the same $ x $-coordinate, show that $ MS $ is parallel to one of the asymptotes of the hyperbola.

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Answer

To show that $MS$ is parallel to one of the asymptotes, we analyze the slopes of the asymptotes as well as the slope between points $M$ and $S$ . If these slopes match, it shows that $MS$ is indeed parallel to one of the asymptotes of the hyperbola.

Using the substitution $ t = \tan \frac{x}{2} $, otherwise, evaluate

Find the volume of the solid.

Show that $ M $ lies on the hyperbola.

Prove that the line through $ P $ and $ Q $ is a tangent to the hyperbola at $ M $.

Show that $ OP \times OQ = \sqrt{a^2t^2 + b^2t^2} \times \sqrt{\frac{a^2}{t^2} + \frac{b^2}{t^2}}$.

If $ P $ and $ S $ have the same $ x $-coordinate, show that $ MS $ is parallel to one of the asymptotes of the hyperbola.

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