Substituting u into a sin equation, Integration, Answer doesn't explain it.

JohnCurrie

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Jul 15, 2016
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Dear Tutors,

I am trying to anti derive formulas for my maths assignment and I didn't understand how to solve one of the questions. Naturally, I checked the answers on an online calculator in order to find what I did wrong and their answer jumps ahead a step without explaining what its working was.

Original Equation: y = 2x ⋅ sin(0.1 ⋅ x^2)

Simpify = 2x ⋅ sin(x^2/10)

substitute u into sin: x^2/10 --> du/dx --> x/5

therefore = 10 ∫sin(u) ??

I don't understand where the 10 came from or why the 2x disappeared.

any help is appreciated, Thank you.
 
Dear Tutors,

I am trying to anti derive formulas for my maths assignment and I didn't understand how to solve one of the questions. Naturally, I checked the answers on an online calculator in order to find what I did wrong and their answer jumps ahead a step without explaining what its working was.

Original Equation: y = 2x ⋅ sin(0.1 ⋅ x^2)

Simpify = 2x ⋅ sin(x^2/10)

substitute u into sin: x^2/10 --> du/dx --> x/5

therefore = 10 ∫sin(u) ??

I don't understand where the 10 came from or why the 2x disappeared.

any help is appreciated, Thank you.


\(\displaystyle \int \mathrm{2x\sin{(0.1x^2)}} \mathrm{d}x =\begin{bmatrix} \dfrac{1}{10}x^2=u \\ \dfrac{1}{5}xdx=du \end{bmatrix}= \\ \int \mathrm{10\sin{(u)}} \mathrm{d}u \)

\(\displaystyle u'dx=du \)

If you plug in substitutes, you get:

\(\displaystyle \int \mathrm{10 \cdot \sin{(\dfrac{1}{10}x^2)}} \cdot \mathrm{\dfrac{1}{5}x} \mathrm{d}x= \int \mathrm{2x\sin{(\dfrac{1}{10}x^2)}} \mathrm{d}x \)

Do you see it now?

You need to put 10 in there so that expression remains equivalent to the beginning expression. And 2x didn't disappear it's contained in \(\displaystyle 10 \cdot \mathrm{d}u = 10 \cdot \mathrm{\dfrac{1}{5}x} \mathrm{d}x \) .
 
Last edited:
Yeah i kind of get it, I'll show this to my teacher on Monday to walk me through it. Thank you :)
 
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