Question

2cos²0 + 5sin³0 +2 differenction and intergtion​

Answer 1

Answer:

function to be differentiable at any point x = a, in its domain, it must be continuous at that particular point but vice-versa is necessarily not always true. The domain of f’(x) is defined by the existence of its limits.

If y = f(x) is a function in x, then the derivative of f(x) is given as dy/dx. This is known as the derivative of y with respect to x.

Also, the derivative of a function f(x) at x = a, is given by:

Answer 2

${\orange{ \boxed{ \boxed{ \begin{array}{cc} \bf \to \: let \\ \\ \rm \: f(\theta) = 2cos {}^{2} \theta + 5sin {}^{2}\theta + 2 \\ \\ \rm = 2 {cos}^{2} \theta + 5(1 - {cos}^{2} \theta) + 2 \\ \\ \rm = 2 {cos}^{2} \theta + 5 - 5cos {}^{2}\theta + 2 \\ \\ \rm = 7 - 3 {cos}^{2} \theta \: \\ \\ \therefore \rm \: f(\theta) = 7 - 3 {cos}^{2} \theta \\ \\ \\ \blue{ \underline{\sf \: we \: have \: to \: find \: : }} \\ \\ \sf \hookrightarrow \: differentiate \: the \: function \: = \frac{df(\theta)}{d\theta \: } \\ \\ \\ \blue{ \sf \hookrightarrow \: integrate \: the \: function = \displaystyle \int \rm \: f(\theta) \: d\theta \: } \end{array}}}}}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \red{ \underline{ \sf \: solution \: : }}$

${\orange{ \boxed{ \boxed{ \begin{array}{cc} \underline{\bf \: differentiation} \\ \\ \rm\frac{df(\theta)}{d\theta} = \frac{d}{d\theta} (7 - 3 {cos}^{2} \theta ) \\ \\ {\pink{ { \boxed{ \begin{array}{cc} \sf \: we \: know \: that : \\ \\ \sf \hookrightarrow \: \frac{d}{dx}(u \pm \: v) = \frac{d \: u}{dx} \pm \: \frac{dv}{dx} \\ \\ \sf \hookrightarrow \: \frac{d}{dx}(constant) = 0 \\ \\ \sf \hookrightarrow \: \frac{d}{dx} cos \: x = - sin \: x \\ \\ \sf \hookrightarrow \: \frac{d}{dx} {x}^{n} = n {x}^{n - 1} \end{array}}}}} \\ \\ \rm = \frac{d}{d\theta}7 - \frac{d}{d\theta} 3 {cos}^{2}\theta \\ \\ \rm = 0 - 3 \frac{ d }{d\theta} {(cos \: \theta)}^{2} \\ \\ \rm = - 3 \times 2cos \: \theta \: \frac{d}{d\theta} \: cos \: \theta \\ \\ \rm = - 3 \times 2cos \: \theta( - sin \: \theta) \\ \\ \rm = 3 \times 2sin\theta \: cos\theta \\ \\ \rm = 3sin2\theta \\ \\ \pink{ \{ \because\sf \: 2sinx \: cos \: x = sin2x \}} \\ \\ \\ \boxed{\therefore \rm \: \frac{df(\theta)}{dx} = 3 \: sin2\theta} \end{array}}}}}$

${\blue{ \boxed{ \boxed{ \begin{array}{cc} \underline{ \bf \: integration} \\ \\ \bf \: let \\ \\ \rm \: I = \displaystyle \int \: \rm \: f(\theta) \: d\theta \\ \\ \rm = \displaystyle \int \rm \:(7 - 3cos {}^{2} \theta) \: \: d\theta \\ \\ \small{{\pink{ { \boxed{ \begin{array}{cc} \sf \: we \: know \: that : \\ \\ \sf \hookrightarrow \: \displaystyle \int \: \rm \: (u \pm \: v) \: dx = \displaystyle \int \: \rm \: u \: dx \pm \: \displaystyle \int \: \rm \: v \: dx \\ \\ \sf \hookrightarrow \: \displaystyle \int \rm \:cos \: x \: dx = sin \: x + c \\ \\ \sf \hookrightarrow \: \displaystyle \int \rm \:k \: dx = kx + c \end{array}}}}} }\\ \\ \rm = \displaystyle \int \rm \:7 \: d\theta - 3\displaystyle \int \: \rm \: {cos}^{2} \theta \: d\theta \\ \\ \rm = 7\theta -3\displaystyle \int \rm \: \frac{1}{2} \times 2cos {}^{2} \theta \: d\theta \\ \\ \rm = 7\theta - \frac{3}{2} \displaystyle \int \: \rm \: (1 + cos2\theta) \: d\theta \\ \\ \rm = 7\theta - \frac{3}{2} \{\displaystyle \int \: \rm \: 1. \: d\theta + \displaystyle \int \: \rm \: cos2\theta \: d\theta \} \\ \\ \rm = 7\theta - \frac{3}{2} \{\theta + \frac{sin2\theta}{2} \} + c \\ \\ \rm = 7\theta - \frac{3\theta}{2} - \frac{3 \: sin2\theta}{4} + c \\ \\ \rm = \frac{11\theta}{2} -\frac{3sin2\theta}{4} + c \\ \\ \rm \: = answer \end{array}}}}}$