# Differentiation and Integration of Trigonometric Functions

The derivative and primitive function rules in Calculus are applicable to Trigonometric Functions. Several rules are listed below:

Simple differentiation and integration of trigonometric functions:

General formula of the derivatives of trigonometric functions:

General formula of the primitives of trigonometric functions:

### Examples

1. Differentiate $\cos{ 5x }$.

$\cfrac { d }{ dx } [\cos { (5x) } ]=-5\sin { (5x) }$

2. Find the exact value of the gradient of the tangent to the curve $y={x}^{2}\sin{x}$ at the point where $x=\cfrac{\pi}{4}$.

$\cfrac { dy }{ dx } ={ u }^{ ' }v+{ v }^{ ' }u\\ \\ \cfrac { dy }{ dx } =\cos { x } ({ x }^{ 2 })+2x\sin { (x) } \\ \\ \cfrac { dy }{ dx } ={ x }^{ 2 }\cos { x } +2x\sin { x } \\ \\ \\ When\quad x=\cfrac { \pi }{ 4 } ,\\ \\ \cfrac { dy }{ dx } ={ \left( \cfrac { \pi }{ 4 } \right) }^{ 2 }\cos { \cfrac { \pi }{ 4 } } +2\left( \cfrac { \pi }{ 4 } \right) \sin { \cfrac { \pi }{ 4 } } \\ \\ \cfrac { dy }{ dx } =\cfrac { { \pi }^{ 2 } }{ 16 } \times \cfrac { 1 }{ \sqrt { 2 } } +\cfrac { \pi }{ 2 } \times \cfrac { 1 }{ \sqrt { 2 } } \\ \\ \cfrac { dy }{ dx } =\cfrac { { \pi }^{ 2 } }{ 16\sqrt { 2 } } +\cfrac { \pi }{ 2\sqrt { 2 } } \\ \\ \cfrac { dy }{ dx } =\cfrac { \pi \sqrt { 2 } (\pi +8) }{ 32 }$

3. Evaluate $\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \cos { x } } dx$.

$=\quad { \left[ \sin { x } \right] }_{ 0 }^{ \frac { \pi }{ 2 } }\\ \\ =\quad \sin { \frac { \pi }{ 2 } } -\sin { 0 } \\ \\ =\quad 1-0\\ =\quad 1$

4. Find the area enclosed between the curve $y=\sin{ x }$, the x-axis, in the domain of $\cfrac{ \pi }{2}$ and $\cfrac{ 3 \pi }{2}$.

The graph above shows the intended area of focus. Remember that, from integration principles, the area below the x-axis would be zero, so we need to use two separate integrations. First, we evaluate the area between $\cfrac{\pi}{2}$ and $\pi$, then we evaluate the negative area between $\cfrac{\pi}{2}$ and $\pi$.

Therefore,

$\int _{ \frac { \pi }{ 2 } }^{ \pi }{ \sin { x } } \quad +\quad \left( -\int _{ \pi }^{ \frac { 3\pi }{ 2 } }{ \sin { x } } \right) \\ \\ \\ =\quad { \left[ -\cos { x } \right] }_{ \frac { \pi }{ 2 } }^{ \pi }\quad -\quad { \left[ -\cos { x } \right] }_{ \pi }^{ \frac { 3\pi }{ 2 } }\\ \\ \\ =\quad \left[ -\cos { \pi } --\cos { \frac { \pi }{ 2 } } \right] \quad -\quad \left[ -\cos { \frac { 3\pi }{ 2 } } --\cos { \pi } \right] \\ \\ =\quad \left[ -(-1)+0 \right] \quad -\quad \left[ 0+-1 \right] \\ \\ =\quad 1\quad -\quad -1\\ \\ =\quad 2\quad { units }^{ 2 }$

5. Find the primitive function of ${ \sin ^{ 6 }{ x } \cos ^{ 3 }{ x } }$ using substitution.