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Math
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2 \times 3
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234
a_{i-1}
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\frac{2}{3}
32
\sqrt{2}
2
\sum_{i=1}^3
∑i=13
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\boxed{123}
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Comments
With cos2A+sin2A=1 and sin(2A)=2sinAcosA, you can simplify the integrand to 4−sin2(2x)2. Then multiply top and bottom by 2, substitute 2sin2(2x)=1−cos(4x), then apply Tangent half angle substitution.
The numerator of the integrand can be factored into (cosx+sinx)(1−cosxsinx). The denominator can be simplified like the previous one, then multiply top and bottom by 2, you're left with 4−sin2(2x)(cosx+sinx)(2−2sinxcosx)=(2−sin(2x))(2+sin(2x))(cosx+sinx)(2−sin(2x))=2+sin(2x)cosx+sinx=3−(sinx−cosx)2cosx+sinx. Set y=sinx−cosx, then you're left with a familiar integral ∫a2−z2dz, in this case a=3. Finish it off via partial fraction - cover up rule, and substitute everything back.
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# I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world"\(...\)or\[...\]to ensure proper formatting.2 \times 32^{34}a_{i-1}\frac{2}{3}\sqrt{2}\sum_{i=1}^3\sin \theta\boxed{123}Comments
With cos2A+sin2A=1 and sin(2A)=2sinAcosA, you can simplify the integrand to 4−sin2(2x)2. Then multiply top and bottom by 2, substitute 2sin2(2x)=1−cos(4x), then apply Tangent half angle substitution.
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Nice !
∫1+sin4(x)+cos4(x)cos3(x)+sin3(x)dx
What about this ?
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The numerator of the integrand can be factored into (cosx+sinx)(1−cosxsinx). The denominator can be simplified like the previous one, then multiply top and bottom by 2, you're left with 4−sin2(2x)(cosx+sinx)(2−2sinxcosx)=(2−sin(2x))(2+sin(2x))(cosx+sinx)(2−sin(2x))=2+sin(2x)cosx+sinx=3−(sinx−cosx)2cosx+sinx. Set y=sinx−cosx, then you're left with a familiar integral ∫a2−z2dz, in this case a=3. Finish it off via partial fraction - cover up rule, and substitute everything back.
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You are a real brilliant! :-)