Integrate Trigonometric Terms

General terms

An example: \begin{align} \int \sec \left(x \over 3\right) \phantom{.} dx & = 3 \int {1 \over 3} \sec \left({1 \over 3}x \right) \phantom{.} dx \\ & = 3 \left[ \ln \left| \sec {x \over 3} + \tan {x \over 3} \right| \right] \\ & = 3 \ln \left| \sec {x \over 3} + \tan {x \over 3} \right| + C \end{align}

Integrating terms in the form of $ f'(x) [f(x)]^n $ & ${f'(x) \over f(x)}$

For $n \ne -1$, $$ \boxed{ \int f'(x) [f(x)]^n \phantom{.} dx = { [f(x)]^{n + 1} \over n + 1} + C } $$

For $n = -1$, $$ \boxed{ \int f'(x) [f(x)]^{-1} \phantom{.} dx = \int {f'(x) \over f(x)} \phantom{.} dx = \ln \left| f(x) \right| + C } $$

An example using the first formula: \begin{align} \int \cos 3x \sin^3 3x & = {1 \over 3} \int 3 \cos 3x (\sin 3x)^3 \phantom{.} dx \\ & = {1 \over 3} \left[ (\sin 3x)^4 \over 4 \right] \\ & = {1 \over 12} \sin^4 3x + C \end{align}

An example using the second formula: \begin{align} \int { \sec^2 {1 \over 2}x \over \tan {1 \over 2}x } \phantom{.} dx & = 2 \int { {1 \over 2} \sec^2 {1 \over 2}x \over \tan {1 \over 2}x } \phantom{.} dx \\ & = 2 \ln \left| \tan {1 \over 2} x \right| + C \end{align}

Integrate sin A cos A, sin² A and cos²A

\begin{align} & \boxed{ \sin 2A = 2\sin A \cos A } \\ \\ & \boxed{ \cos 2A = \cos^2 A - \sin^2 A = 2\cos^2 A - 1 = 1 - 2\sin^2 A} \end{align}

Integrate sin A cos A

Use the formula sin 2A = 2 sin A cos A: \begin{align} \int \sin 3x \cos 3x \phantom{.} dx & = {1 \over 2} \int 2 \sin 3x \cos 3x \phantom{.} dx \\ & = {1 \over 2} \int \sin 6x \phantom{.} dx \\ & = {1 \over 12} \int 6 \sin 6x \phantom{.} dx \phantom{00000} \left[ \int f'(x) \sin [f(x)] \phantom{.} dx \right]\\ & = {1 \over 12} (-\cos 6x) \\ & = -{1 \over 12} \cos 6x + C \end{align}

Integrate sin²A

From cos 2A = 1 - 2sin² A, make sin² A the subject: $$ \sin^2 A = {1 \over 2} (1 - \cos 2A) = {1 \over 2} - {1 \over 2}\cos 2A $$

Applying the formula: \begin{align} \int \sin^2 2x \phantom{.} dx & = \int \left( {1 \over 2} - {1 \over 2} \cos 4x \right) \phantom{.} dx \\ & = \int {1 \over 2} \phantom{.} dx - {1 \over 2} \int \cos 4x \phantom{.} dx \\ & = {1 \over 2}x - {1 \over 8} \int 4 \cos 4x \phantom{.} dx \\ & = {1 \over 2}x - {1 \over 8} \sin 4x + C \end{align}

Integrate cos²A

From cos 2A = 2cos² A - 1, make cos² A the subject: $$ \cos^2 A = {1 \over 2}(\cos 2A + 1) = {1 \over 2} \cos 2A + {1 \over 2} $$

Applying the formula: \begin{align} \int 2 \cos^2 \left(x \over 3\right) \phantom{.} dx & = \int 2 \left[ {1 \over 2} \cos \left(2x \over 3\right) + {1 \over 2} \right] \phantom{.} dx \\ & = \int \left[ \cos \left({2 \over 3}x\right) + 1 \right] \phantom{.} dx \\ & = \int \cos \left( {2 \over 3}x \right) \phantom{.} dx + \int 1 \phantom{.} dx \\ & = {3 \over 2} \int {2 \over 3} \cos \left( {2 \over 3}x \right) \phantom{.} dx + x \\ & = {3 \over 2} \sin \left( {2 \over 3}x \right) + x + C \end{align}

Integrate tan² A and cot²A

Use the following 2 identities learnt in O Level A Maths (not in MF26):

$$ \boxed{ \sec^2 A = 1 + \tan^2 A } $$

$$ \boxed{ \text{cosec}^2 A = 1 + \cot^2 A} $$

The last identity is $\sin^2 A + \cos^2 A = 1$.

Integrate tan²A

From sec²A = 1 + tan² A, make tan² A the subject: $$ \tan^2 A = \sec^2 A - 1 $$

Applying the formula: \begin{align} \int 2 \tan^2 {x \over 3} \phantom{.} dx & = \int 2 \left( \sec^2 {x \over 3} - 1 \right) \phantom{.} dx \\ & = \int 2 \sec^2 {x \over 3} \phantom{.} dx - \int 2 \phantom{.} dx \\ & = 6 \int {1 \over 3} \sec^2 {x \over 3} \phantom{.} dx - 2x \\ & = 6 \tan {x \over 3} - 2x + C \end{align}

Integrate cot²A

From cosec²A = 1 + cot² A, make cot² A the subject: $$ \cot^2 A = \text{cosec}^2 A - 1 $$

Applying the formula: \begin{align} \int \cot^2 (\pi x) \phantom{.} dx & = \int \text{cosec}^2 (\pi x) - 1 \phantom{.} dx \\ & = \int \text{cosec}^2 (\pi x) \phantom{.} dx - \int 1 \phantom{.} dx \\ & = {1 \over \pi} \int \pi \text{ cosec}^2 (\pi x) \phantom{.} dx - x \\ & = {1 \over \pi} [ -\cot (\pi x) ] - x \\ & = -{1 \over \pi} \cot (\pi x) - x + C \end{align}

Simplify by Factor formula

\begin{align} & \boxed{ \sin P + \sin Q \equiv 2 \sin {1 \over 2}(P + Q) \cos {1 \over 2}(P - Q) } \\ & \boxed{ \sin P - \sin Q \equiv 2 \cos {1 \over 2}(P + Q) \sin {1 \over 2}(P - Q) } \\ & \boxed{ \cos P + \cos Q \equiv 2 \cos {1 \over 2}(P + Q) \cos {1 \over 2}(P - Q) } \\ & \boxed{ \cos P - \cos Q \equiv - 2 \sin {1 \over 2}(P + Q) \sin {1 \over 2}(P - Q)} \end{align}

We use the formulas above for expressions in the form of:

1. $\sin ax \cos bx $ (use first or second formula)
2. $\cos ax \cos bx $ (use third formula)
3. $\sin ax \sin bx $ (use fourth formula)

Example

To find the $\int \sin 4\theta \sin 2\theta \phantom{.} d\theta$, manipulate the expression using the fourth formula: $$ \text{Let } \sin 4 \theta \sin 2\theta = \sin {1 \over 2}(P + Q) \theta \sin {1 \over 2}(P - Q) \theta $$

\begin{align} 4 & = {1 \over 2}(P + Q) & & & 2 & = {1 \over 2}(P - Q) \\ 8 & = P + Q \text{ --- (1)} & & & 4 & = P - Q \text{ --- (2)} \\ \\ (1) & + (2), \\ 12 & = 2P \\ 6 & = P \\ \\ 8 & = 6 + Q \\ 2 & = Q \end{align} \begin{align} \text{Since } \cos P - \cos Q & = -2\sin {1 \over 2}(P + Q) \sin {1 \over 2}(P - Q), \\ \\ \cos 6\theta - \cos 2\theta & = -2\sin 4\theta \sin 2 \theta \\ -{1 \over 2}(\cos 6\theta - \cos 2\theta) & = \sin 4\theta \sin 2\theta \end{align}

Now find $\int \sin 4\theta \sin 2\theta \phantom{.} d\theta $: \begin{align} \int \sin 4\theta \sin 2\theta \phantom{.} d\theta & = \int -{1 \over 2}(\cos 6\theta - \cos 2\theta) \phantom{.} d \theta \\ & = -{1 \over 2} \int \cos 6 \theta \phantom{.} d\theta + {1 \over 2} \int \cos 2 \theta \phantom{.} d \theta \\ & = -{1 \over 12} \int 6 \cos 6 \theta \phantom{.} d\theta + {1 \over 4} \int 2 \cos 2\theta \phantom{.} d\theta \\ & = -{1 \over 12} \sin 6\theta + {1 \over 4} \sin 2\theta + C \end{align}