09Cintbyrecognition


 * Integration by Recognition **

Some functions can be difficult to integrate directly, but the antiderivative can be found by recognising the function as being a derivative we have already found.


 * NOTE: ** When part (b) of a question starts with the word ** __hence__ **, you are expected to use the result from part (a). You will not gain marks by using some other method to solve the problem.


 * Example 1 **

a) Find the derivative of y = (5x + 1) 3

Using the chain rule math \\ . \qquad \dfrac{dy}{dx} = 3(5x+1)^2(5) \qquad. \\ . \\ . \qquad \dfrac{dy}{dx} = 15(5x+1)^2 math

b) __Hence__ find the antiderivative of 30(5x + 1) 2

from (a) we know that: math . \qquad \int 15(5x+1)^2 \, dx = (5x+1)^3 + c \qquad. math

Therefore (manipulate to fit the question) math . \qquad \frac{1}{2} \int 30(5x+1)^2 \, dx = (5x+1)^3 + c \qquad. math

Hence math . \qquad \int 30(5x+1)^2 \, dx = 2(5x+1)^3 + c_1 \qquad. math


 * Note:** In my final answer I could have written + 2c, but c is any constant so 2c is also any constant. I simply replace 2c with a new constant c 1.


 * Example 2 **

a) Find the derivative of y = xcos(x)

Using the product rule math \\ . \qquad \dfrac{dy}{dx}= \big( 1 \big) \big( \cos (x) \big) + \big( x \big) \big(-\sin (x) \big) \qquad. \\ . \\ . \qquad \dfrac{dy}{dx} = \cos(x) - x \sin (x) math

b) __Hence__ find the antiderivative of xsin(x)

from (a) we know that: math . \qquad \int \cos(x) - x\sin(x) \; dx = x \cos(x) + c \qquad. math

Therefore math . \qquad \int \cos(x) \, dx - \int x \sin(x) \, dx = x \cos(x) + c \qquad. math

We can integrate the first term: math . \qquad \sin(x) - \int x \sin(x) \, dx = x \cos(x) + c \qquad. math

Hence math \\ . \qquad - \int x \sin(x) \, dx = -\sin(x) + x \cos(x) + c \qquad. \\ . \\ . \qquad \int x \sin(x) \, dx = \sin(x) - x \cos(x) + c_1 math


 * Note**: Again in my final answer I have replaced the constant -c with a new constant c 1.