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Math
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Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.
2 \times 3
2×3
2^{34}
234
a_{i-1}
ai−1
\frac{2}{3}
32
\sqrt{2}
2
\sum_{i=1}^3
∑i=13
\sin \theta
sinθ
\boxed{123}
123
Comments
By Differentiation - Quotient Rule , the integrand might be in the form of v2vu′−uv′, so we can make the assumption that v=xsin(x)+cos(x).
Now we just want to find the unknown coefficients for u=Axcos(x)+Bxsin(x)+Csin(x)+Dcos(x)+Ex such that vu′−uv′=x2. Solving this gives u=sin(x)−xcos(x).
Checking back, the indefinite integral of xsin(x)+cos(x)x2 is indeed xsin(x)+cos(x)sin(x)−xcos(x). Substituting the limits give the desired answer.
Easy Math Editor
This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.
When posting on Brilliant:
*italics*
or_italics_
**bold**
or__bold__
paragraph 1
paragraph 2
[example link](https://brilliant.org)
> This is a quote
\(
...\)
or\[
...\]
to ensure proper formatting.2 \times 3
2^{34}
a_{i-1}
\frac{2}{3}
\sqrt{2}
\sum_{i=1}^3
\sin \theta
\boxed{123}
Comments
By Differentiation - Quotient Rule , the integrand might be in the form of v2vu′−uv′, so we can make the assumption that v=xsin(x)+cos(x).
Now we just want to find the unknown coefficients for u=Axcos(x)+Bxsin(x)+Csin(x)+Dcos(x)+Ex such that vu′−uv′=x2. Solving this gives u=sin(x)−xcos(x).
Checking back, the indefinite integral of xsin(x)+cos(x)x2 is indeed xsin(x)+cos(x)sin(x)−xcos(x). Substituting the limits give the desired answer.
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Challenge student note: nice solution. U have used the trick perfectly.
Yes,thank you!Actually this problem is from a book,the book too had a similar solution,I just wanted a different one,an easier one!
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Please don't use square brackets if it is not a gif. It confuses.
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I will post this one later. Got it.