(a)

\begin{align} \int \sin x + 2 \phantom{.} dx & = {-\cos x \over 1} + 2x \\ & = -\cos x + 2x + c \end{align}

(b)

\begin{align} \int 1 - 3\cos x \phantom{.} dx & = x - {3\sin x \over 1} \\ & = x - 3\sin x + c \end{align}

(c)

\begin{align} \int 5\sec^2 x + 3 \phantom{.} dx & = {5\tan x \over 1} + 3x \\ & = 5\tan x + 3x + c \end{align}

(a)

\begin{align} \int_0^{\pi \over 6} \cos x \phantom{.} dx & = \left[ \sin x \over 1 \right]_0^{\pi \over 6} \\ & = [ \sin x]_0^{\pi \over 6} \\ & = \sin {\pi \over 6} - \sin 0 \\ & = {1 \over 2} - 0 \\ & = {1 \over 2} \end{align}

(b)

\begin{align} \int_0^{\pi \over 4} \sec^2 x \phantom{.} dx & = \left[ \tan x \over 1\right]_0^{\pi \over 4} \\ & = [\tan x]_0^{\pi \over 4} \\ & = \tan {\pi \over 4} - \tan 0 \\ & = 1 - 0 \\ & = 1 \end{align}

(c)

\begin{align} \int_0^{\pi \over 2} \sin x \phantom{.} dx & = \left[ -\cos x \over 1 \right]_0^{\pi \over 2} \\ & = [-\cos x]_0^{\pi \over 2} \\ & = -\cos {\pi \over 2} - (- \cos 0) \\ & = - 0 - (- 1) \\ & = 1 \end{align}

(d)

\begin{align} \int_0^{\pi \over 2} 1 - 2\sin x \phantom{.} dx & = \left[ x - {-2\cos x \over 1} \right]_0^{\pi \over 2} \\ & = \left[ x + 2\cos x \right]_0^{\pi \over 2} \\ & = \left[ {\pi \over 2} + 2\cos {\pi \over 2}\right] - \left[ 0 + 2\cos 0 \right] \\ & = \left[ {\pi \over 2} + 2(0)\right] - [0 + 2(1)] \\ & = {\pi \over 2} - 2 \end{align}

(e)

\begin{align} \int_0^{\pi \over 6} 3 \cos x - 2 \phantom{.} dx & = \left[ {3 \sin x \over 1} - 2x \right]_0^{\pi \over 6} \\ & = \left[ 3\sin x - 2x \right]_0^{\pi \over 6} \\ & = \left[3\sin {\pi \over 6} - 2\left(\pi \over 6\right) \right] - \left[ 3\sin 0 - 2(0)\right] \\ & = \left[3\left(1 \over 2\right) - {\pi \over 3} \right] - \left[3(0) - 0\right] \\ & = {3 \over 2} - {\pi \over 3} \end{align}

(a)

\begin{align} \int \cos 2x \phantom{.} dx & = {\sin 2x \over 2} \\ & = {1 \over 2}\sin 2x + c \end{align}

(b)

\begin{align} \int \sin 3x \phantom{.} dx & = {-\cos 3x \over 3}\\ & = -{1 \over 3}\cos 3x + c \end{align}

(c)

\begin{align} \int 2\cos 4x \phantom{.} dx & = {2(\sin 4x) \over 4} \\ & = {\sin 4x \over 2} \\ & = {1 \over 2}\sin 4x + c \end{align}

(d)

\begin{align} \int -3 \sin {1 \over 2}x \phantom{.} dx & = {-3 \left(-\cos {1 \over 2}x\right) \over {1 \over 2} } \\ & = {3 \cos {1 \over 2}x \over {1 \over 2}} \\ & = 6\cos {1 \over 2}x + c \phantom{000000} \left[ 3 \div {1 \over 2} = 6\right] \end{align}

(e)

\begin{align} \int 4 \sec^2 5x \phantom{.} dx & = {4 (\tan 5x) \over 5} \\ & = {4 \over 5} \tan 5x + c \end{align}

(a)

\begin{align} \int 2 \cos \left({x \over 2} + 1\right) \phantom{0} dx & = {2 \left[ \sin \left({x \over 2} + 1\right)\right] \over {1 \over 2}} \\ & = 4\sin \left( {x \over 2} + 1 \right) + c \phantom{000000} \left[ 2 \div {1 \over 2} = 4 \right] \end{align}

(b)

\begin{align} \int 3 \sin (2 - x) \phantom{.} dx & = {-3 [\cos (2 - x)] \over - 1} \\ & = 3 \cos (2 - x) + c \end{align}

(c)

\begin{align} \int 4 \sec^2 \left( 8x - {\pi \over 4} \right) \phantom{0} dx & = {4 \left[ \tan \left( 8x - {\pi \over 4}\right) \right] \over 8} \\ & = {\tan \left( 8x - {\pi \over 4} \right) \over 2} \\ & = {1 \over 2} \tan \left( 8x - {\pi \over 4} \right) + c \end{align}

(a)

\begin{align} \int \sec^2 x - 4\sin x \phantom{.} dx & = {\tan x \over 1} - {4 (-\cos x) \over 1} \\ & = \tan x + 4\cos x + c \end{align}

(b)

\begin{align} \int 3 \cos x - 2\sin x \phantom{.} dx & = {3 (\sin x) \over 1} - {-2(\cos x) \over 1} + c \\ & = 3 \sin x + 2\cos x + c \end{align}

(c)

\begin{align} \int 4\cos x + 3\sec^2 x \phantom{.} dx & = {4 (\sin x) \over 1} + {3 (\tan x) \over 1} + c \\ & = 4\sin x + 3\tan x + c \end{align}

\begin{align} \int_0^{\pi \over 4} 1 + \tan^2 x \phantom{.} dx & = \int_0^{\pi \over 4} \sec^2 x \phantom{.} dx \phantom{0000000000} [\text{Identity: } 1 + \tan^2 A = \sec^2 A] \\ & = \left[ \tan x \right]_0^{\pi \over 4} \\ & = \tan {\pi \over 4} - \tan 0 \\ & = 1 - 0 \\ & = 1 \end{align}

\begin{align} \int_0^{\pi \over 4} {1 + \cos^2 x \over \cos^2 x} \phantom{.} dx & = \int_0^{\pi \over 4} {1 \over \cos^2 x} + {\cos^2 x \over \cos^2 x} \phantom{.} dx \\ \\ & = \int_0^{\pi \over 4} \sec^2 x + 1 \phantom{.} dx \\ \\ & = \left[ {\tan x \over 1} + x \right]_0^{\pi \over 4} \\ \\ & = [\tan x + x ]_0^{\pi \over 4} \\ \\ & = \left[ \tan {\pi \over 4} + {\pi \over 4} \right] - \left[ \tan 0 + 0\right] \\ \\ & = [1 + {\pi \over 4}] - [0 + 0] \\ \\ & = 1 + {\pi \over 4} \phantom{00} \text{ (Shown)} \end{align}

(a)

\begin{align} \int -\sin (2x + 1) \phantom{.} dx & = -{-\cos (2x + 1) \over 2} \\ & = {\cos (2x + 1) \over 2} \\ & = {1 \over 2}\cos (2x + 1) + c \end{align}

(b)

\begin{align} \int \cos \left(2x + {\pi \over 4}\right) \phantom{.} dx & = {\sin \left(2x + {\pi \over 4}\right) \over 2} \\ & = {1 \over 2}\sin \left(2x + {\pi \over 4}\right) + c \end{align}

(c)

\begin{align} \int 6\sin (3x + 2) \phantom{.} dx & = {6 [-\cos (3x + 2)] \over 3} \\ & = -2\cos (3x + 2) + c \end{align}

(a)

\begin{align} \int_{\pi \over 12}^{\pi \over 3} \cos \left(2x + {\pi \over 3}\right) \phantom{.} dx & = \left[ {\sin \left(2x + {\pi \over 3}\right) \over 2} \right]_{\pi \over 12}^{\pi \over 3} \\ & = {\sin \left(2({\pi \over 3}) + {\pi \over 3}\right) \over 2} - {\sin \left(2({\pi \over 12}) + {\pi \over 3}\right) \over 2} \\ & = {\sin \left( {2\pi \over 3} + {\pi \over 3} \right) \over 2} - {\sin \left( {\pi \over 6} + {\pi \over 3} \right) \over 2} \\ & = {\sin \pi \over 2} - {\sin {\pi \over 2} \over 2} \\ & = {0 \over 2} - {1 \over 2} \\ & = -{1 \over 2} \end{align}

(b)

\begin{align} \int_{\pi \over 2}^{\pi} 2 \sin (\pi - x) \phantom{0} dx & = \left[ {2[-\cos (\pi - x)] \over -1} \right]_{\pi \over 2}^{\pi} \\ & = \left[ 2 \cos (\pi - x) \right]_{\pi \over 2}^{\pi} \\ & = 2 \cos (\pi - \pi) - 2 \cos \left(\pi - {\pi \over 2}\right) \\ & = 2 \cos 0 - 2 \cos {\pi \over 2} \\ & = 2 (1) - 2(0) \\ & = 2 \end{align}

(c) The special values used are $\sin {\pi \over 6} = \sin 30^\circ = {1 \over 2}$ and $\sin {\pi \over 3} = \sin 60^\circ = {\sqrt{3} \over 2}$.

\begin{align} \int_0^1 x - \cos \left({\pi \over 3} - {\pi \over 6}x\right) \phantom{.} dx & = \left[ {x^2 \over 2} - {\sin \left({\pi \over 3} - {\pi \over 6}x\right) \over -{\pi \over 6}} \right]_0^1 \\ \\ & = \left[ {1 \over 2}x^2 + {6 \over \pi} \sin \left({\pi \over 3} - {\pi \over 6}x \right) \right]_0^1 \\ \\ & = \left[ {1 \over 2}(1)^2 + {6 \over \pi} \sin \left({\pi \over 3} - {\pi \over 6}(1) \right) \right] - \left[{1 \over 2}(0)^2 + {6 \over \pi} \sin \left({\pi \over 3} - {\pi \over 6}(0) \right) \right] \\ \\ & = \left[ {1 \over 2} + {6 \over \pi} \sin {\pi \over 6} \right] - \left[ 0 + {6 \over \pi} \sin {\pi \over 3} \right] \\ \\ & = \left[ {1 \over 2} + {6 \over \pi} \left(1 \over 2\right) \right] - \left[ {6 \over \pi} \left( \sqrt{3} \over 2\right) \right] \\ \\ & = {1 \over 2} + {3 \over \pi} - {3\sqrt{3} \over \pi} \\ \\ & = {1 \over 2} + {3 - 3\sqrt{3} \over \pi} \end{align}

(d)

\begin{align} \int_0^\pi \sin 5t \phantom{0} dt & = \left[ {-\cos 5t \over 5} \right]_0^\pi \\ & = \left[-{1 \over 5} \cos 5t \right]_0^\pi \\ & = \left[ -{1 \over 5} \cos \left(5(\pi)\right) \right] - \left[ -{1 \over 5} \cos \left(5(0)\right) \right] \\ & = \left[ -{1 \over 5} \cos 5\pi \right] - \left[ -{1 \over 5} \cos 0 \right] \\ & = \left[ -{1 \over 5} ( - 1) \right] - \left[ -{1 \over 5} (1) \right] \\ & = {1 \over 5} + {1 \over 5} \\ & = {2 \over 5} \end{align}

(i)

\begin{align} u & = x &&& v & = \cos x \\ {du \over dx} & = 1 &&& {dv \over dx} & = - \sin x \end{align} \begin{align} {d \over dx} (x \cos x) & = (x)(- \sin x) + (\cos x)(1) \phantom{000000} [\text{Product rule}] \\ & = - x \sin x + \cos x \end{align}

(ii)

\begin{align} {d \over dx} (x \cos x) & = - x \sin x + \cos x \\ \\ \implies \int - x \sin x + \cos x \phantom{.} dx & = x \cos x \\ \\ \int - x \sin x \phantom{.} dx + \int \cos x \phantom{.} dx & = x \cos x \\ \int - x \sin x \phantom{.} dx & = x \cos x - \int \cos x \phantom{.} dx \\ - \int x \sin x \phantom{.} dx & = x \cos x - \sin x \\ \int x \sin x \phantom{.} dx & = - x \cos x + \sin x + c \end{align}

\begin{align} \int (\sin x - \cos x)^2 \phantom{.} dx & = \int \sin^2 x - 2\sin x \cos x + \cos^2 x \phantom{.} dx \\ & = \int \sin^2 x + \cos^2 x - 2\sin x \cos x \phantom{0} dx \\ & = \int 1 - 2\sin x \cos x \phantom{.} dx \phantom{00000000000} [\text{Identity: } \sin^2 A + \cos^2 A = 1 ] \\ & = \int 1 - \sin 2x \phantom{.} dx \phantom{00000000000000.} [\text{Double angle formula: } \sin 2A = 2\sin A \cos A ] \\ & = x - {- \cos 2x \over 2} + c \\ & = x + {1 \over 2} \cos 2x + c \\ & = x + {1 \over 2} (1 - 2\sin^2 x) + c \phantom{000000000(.} [\text{Double angle formula: } \cos 2A = 1 - 2\sin^2 A] \\ & = x + {1 \over 2} - \sin^2 x + c \\ & = x - \sin^2 x + {1 \over 2} + c \\ & = x - \sin^2 x + c_1 \text{, where } c_1 \text{ is an arbitrary constant and } c_1 = {1 \over 2} + c \phantom{00} \text{ (Shown)} \end{align}

(i)

\begin{align} {d \over dx} \left[ \sin (x^2 + 2x) \right] & = (2x +2) \left[ \cos (x^2 + 2x)\right] \\ & = (2x + 2)\cos (x^2 + 2x) \\ & = 2(x + 1)\cos (x^2 + 2x) \end{align}

(ii)

\begin{align} {d \over dx} \left[ \sin (x^2 + 2x) \right] & = 2(x + 1)\cos (x^2 + 2x) \\ \\ \implies \int 2(x + 1)\cos (x^2 + 2x) \phantom{.} dx & = \sin (x^2 + 2x) \\ \\ 2\int (x + 1)\cos (x^2 + 2x) \phantom{.} dx & = \sin (x^2 + 2x) \\ \int (x + 1)\cos (x^2 + 2x) \phantom{.} dx & = {1 \over 2}\sin (x^2 + 2x) + c \end{align}

\begin{align} \int \sin x \cos x \phantom{.} dx & = {1 \over 2} \int 2 \sin x \cos x \phantom{.} dx \\ & = {1 \over 2} \int \sin 2x \phantom{.} dx \phantom{000000} [\text{Double angle formula: } \sin 2A = 2 \sin A \cos A ] \\ & = {1 \over 2} \left( - \cos 2x \over 2 \right) \\ & = -{1 \over 4} \cos 2x \\ \\ \text{Using } & \cos 2A = 2\cos^2 A - 1, \\ \int \sin x \cos x \phantom{.} dx & = -{1 \over 4} (2 \cos^2 x - 1) \\ & = -{1 \over 2}\cos^2 x + {1 \over 4} + c \\ \\ \therefore \text{Paul } & \text{is correct} \\ \\ \\ \text{Using } & \cos 2A = 1 - 2 \sin^2 A, \\ \int \sin x \cos x \phantom{.} dx & = -{1 \over 4} (1 - 2\sin^2 x) \\ & = -{1 \over 4} + {1 \over 2} \sin^2 x \\ & = {1 \over 2} \sin^2 x - {1 \over 4} + c \\ \\ \therefore \text{Mark } & \text{is correct as well} \end{align}