07 2intbyrec

# Integration by Recognition

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

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Note: In this course, questions of this type will always be in two parts.
a) .. Find the derivative of some function $f(x)$
b) .. Hence, find the antiderivative of some expression $g(x)$
… … where $g(x)$ will be some variation of $f'(x)$.

To solve these questions use the fact that if $f'(x)$ is the derivative of $f(x)$
Then $f(x) = \displaystyle{\int} f'(x) \, dx + c$

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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.

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### Example 1

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

b) .. Hence find the antiderivative of $30(5x + 1)^2$

Solution

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

… … … Using the chain rule

… … … $\dfrac{dy}{dx} = 3(5x+1)^2(5)$

… … …$\dfrac{dy}{dx} = 15(5x+1)^2$
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b) .. Hence find the antiderivative of $30(5x + 1)^2$

… … … from (a) we know that:

… … … $\displaystyle{\int } 15(5x+1)^2 \, dx = (5x+1)^3 + c$
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… … … Therefore (manipulate to fit the question)

… … … $\dfrac{1}{2} \displaystyle{\int} 30(5x+1)^2 \, dx = (5x+1)^3 + c$
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… … … Hence

… … … $\displaystyle{\int} 30(5x+1)^2 \, dx = 2(5x+1)^3 + c_1$
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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 c1.

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### Example 2

a) .. Find the derivative of $y = x\cos(x)$

b) .. Hence find the antiderivative of $x\sin(x)$

Solution

a) .. Find the derivative of $y = x\cos(x)$

… … … Using the product rule

… … … $\dfrac{dy}{dx}= \big( 1 \big) \big( \cos (x) \big) + \big( x \big) \big(-\sin (x) \big)$

… … … $\dfrac{dy}{dx} = \cos(x) - x \sin (x)$
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b) .. Hence find the antiderivative of $x\sin(x)$

… … … from (a) we know that:

… … … $\displaystyle{\int} \cos(x) - x\sin(x) \; dx = x \cos(x) + c$
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… … … Therefore (rearrange to fit the questions)

… … … $\displaystyle{\int} \cos(x) \, dx - \displaystyle{\int} x \sin(x) \, dx = x \cos(x) + c$
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… … … We can integrate the first term:

… … … $\sin(x) - \displaystyle{\int} x \sin(x) \, dx = x \cos(x) + c$
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… … … Hence

… … … $- \displaystyle{\int} x \sin(x) \, dx = -\sin(x) + x \cos(x) + c$

… … … $\displaystyle{\int} x \sin(x) \, dx = \sin(x) - x \cos(x) + c_1$
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Note: Again in my final answer I have replaced the constant -c with a new constant c1.

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