Definite integral with trigonometric functions

3 \includegraphics[max width=\textwidth, alt={}, center]{ed12a4fb-e3bf-4d00-ad09-9ba5be941dd5-04_531_739_258_703} The diagram shows the curve with equation \(y = 3 \sin x - 3 \sin 2 x\) for \(0 \leqslant x \leqslant \pi\). The curve meets the \(x\)-axis at the origin and at the points with \(x\)-coordinates \(a\) and \(\pi\).

1. Find the exact value of \(a\).
2. Find the area of the shaded region.

6 Show that \(\int _ { \frac { 1 } { 4 } \pi } ^ { \frac { 1 } { 3 } \pi } \left( 4 \cos ^ { 2 } 2 x + \frac { 1 } { \cos ^ { 2 } x } \right) \mathrm { d } x = \frac { 3 } { 4 } \sqrt { 3 } + \frac { 1 } { 6 } \pi - 1\).

3 Find the exact value of \(\int _ { 0 } ^ { \frac { 1 } { 6 } \pi } ( \cos 2 x + \sin x ) \mathrm { d } x\).

6 \includegraphics[max width=\textwidth, alt={}, center]{3b217eb4-3bd3-4800-a913-749754bf109f-3_401_586_817_778} The diagram shows part of the curve with equation $$y = 4 \sin ^ { 2 } x + 8 \sin x + 3$$ and its point of intersection \(P\) with the \(x\)-axis.

1. Find the exact \(x\)-coordinate of \(P\).
2. Show that the equation of the curve can be written $$y = 5 + 8 \sin x - 2 \cos 2 x$$ and use integration to find the exact area of the shaded region enclosed by the curve and the axes.

5 \includegraphics[max width=\textwidth, alt={}, center]{4c71f68a-efb9-4408-bf03-874e0d4426d5-2_458_807_1786_667} The diagram shows the curve $$y = 8 \sin \frac { 1 } { 2 } x - \tan \frac { 1 } { 2 } x$$ for \(0 \leqslant x < \pi\). The \(x\)-coordinate of the maximum point is \(\alpha\) and the shaded region is enclosed by the curve and the lines \(x = \alpha\) and \(y = 0\).

1. Show that \(\alpha = \frac { 2 } { 3 } \pi\).
2. Find the exact value of the area of the shaded region.

4

1. Find the exact value of \(\int _ { 0 } ^ { \frac { 1 } { 2 } \pi } \left( 4 \sin 2 x + 2 \cos ^ { 2 } x \right) \mathrm { d } x\). Show all necessary working.
2. Use the trapezium rule with two intervals to find an approximation to \(\int _ { 2 } ^ { 8 } \sqrt { } ( \ln ( 1 + x ) ) \mathrm { d } x\)

10. \begin{figure}[h] \end{figure} Figure 4 shows a sketch of part of the curve with equation $$y = ( 1 + 2 \cos 2 x ) ^ { 2 }$$

1. Show that $$( 1 + 2 \cos 2 x ) ^ { 2 } \equiv p + q \cos 2 x + r \cos 4 x$$ where \(p , q\) and \(r\) are constants to be found. The curve touches the positive \(x\)-axis for the second time when \(x = a\), as shown in Figure 4. The regions bounded by the curve, the \(y\)-axis and the \(x\)-axis up to \(x = a\) are shown shaded in Figure 4.
2. Find, using algebraic integration and making your method clear, the exact total area of the shaded regions. Write your answer in simplest form. \includegraphics[max width=\textwidth, alt={}, center]{9b0b8db0-79fd-4ad5-88c9-737447d9f894-32_2255_51_313_1980}

1 Evaluate \(\int _ { 0 } ^ { \frac { 1 } { 6 } \pi } \cos 3 x \mathrm {~d} x\).

3 Evaluate \(\int _ { 0 } ^ { \frac { 1 } { 6 } \pi } ( 1 - \sin 3 x ) \mathrm { d } x\), giving your answer in exact form.

3 Evaluate \(\int _ { 0 } ^ { \frac { 1 } { 6 } \pi } \sin 3 x \mathrm {~d} x\).
[0pt] [3]

2 Evaluate \(\int _ { 0 } ^ { \frac { 1 } { 6 } \pi } \cos 3 x \mathrm {~d} x\).

1. Evaluate

$$\int _ { 0 } ^ { \pi } \sin x ( 1 + \cos x ) d x$$

5. \begin{figure}[h] \end{figure} Show that the area of the finite region between the curves \(y = \tan ^ { 2 } x\) and \(y = 4 \cos 2 x - 1\) in the interval \(- \frac { \pi } { 2 } < x < \frac { \pi } { 2 }\), shown shaded in Figure 3, is given by $$2 \sqrt { 2 \sqrt { 3 } } - 2 \sqrt { 2 \sqrt { 3 } - 3 }$$

1 Evaluate \(\int _ { 0 } ^ { \frac { 1 } { 6 } \pi } \sin 3 x \mathrm {~d} x\).

4 Find the exact value of \(\int _ { 0 } ^ { \frac { 1 } { 4 } \pi } ( 1 + \sin x ) ^ { 2 } \mathrm {~d} x\).

7

1. Given that \(y = \ln ( 1 + \sin x ) - \ln ( \cos x )\), show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 1 } { \cos x }\).
2. Using this result, evaluate \(\int _ { 0 } ^ { \frac { 1 } { 3 } \pi } \sec x \mathrm {~d} x\), giving your answer as a single logarithm.

7 Find the exact value of \(\int _ { 0 } ^ { \frac { 1 } { 6 } \pi } ( 1 - \sin 3 x ) ^ { 2 } \mathrm {~d} x\).

5

1. Show that \(\frac { 1 } { 1 - \tan x } - \frac { 1 } { 1 + \tan x } \equiv \tan 2 x\).
2. Hence evaluate \(\int _ { \frac { 1 } { 12 } \pi } ^ { \frac { 1 } { 6 } \pi } \left( \frac { 1 } { 1 - \tan x } - \frac { 1 } { 1 + \tan x } \right) \mathrm { d } x\), giving your answer in the form \(a \ln b\).

8

1. Write down \(\int \sec ^ { 2 } x \mathrm {~d} x\).
2. Given that \(y = \frac { \cos x } { \sin x }\), use the quotient rule to show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = - \operatorname { cosec } ^ { 2 } x\).
3. Prove the identity \(( \tan x + \cot x ) ^ { 2 } = \sec ^ { 2 } x + \operatorname { cosec } ^ { 2 } x\).
4. Hence find \(\int _ { 0.5 } ^ { 1 } ( \tan x + \cot x ) ^ { 2 } \mathrm {~d} x\), giving your answer to two significant figures.

1. By differentiating \(\frac{\cos x}{\sin x}\), show that if \(y = \cot x\) then \(\frac{dy}{dx} = -\cosec^2 x\). [3]
2. Hence show that \(\int_{\frac{\pi}{6}}^{\frac{\pi}{2}} \cosec^2 x \, dx = \sqrt{3}\). [2]

By using appropriate trigonometrical identities, find the exact value of

1. \(\int_{\frac{\pi}{6}}^{\frac{\pi}{2}} \cot^2 x \, dx\), [3]
2. \(\int_{\frac{\pi}{6}}^{\frac{\pi}{2}} \frac{1}{1 - \cos 2x} \, dx\). [3]

Evaluate $$\int_0^{\frac{\pi}{4}} \sin 2x \cos x \, dx.$$ [5]

Evaluate \(\int_0^{\frac{\pi}{12}} \cos 3x \, dx\), giving your answer in exact form. [3]