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

[JohnCurrie]

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.

 

[Johulus]

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 $\displaystyle 10 \cdot \mathrm{d}u = 10 \cdot \mathrm{\dfrac{1}{5}x} \mathrm{d}x$ .

 

[JohnCurrie]

Yeah i kind of get it, I'll show this to my teacher on Monday to walk me through it. Thank you