#Definite Integration

54 messages · Page 1 of 1 (latest)

high zenith
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It's basic integration yet I couldn't solve it.

hexed cryptBOT
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steep spire
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let u=pi-x

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then ezpz

high zenith
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OK let me try that

steep spire
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$I=\int_0^{\pi}\frac{(\pi-u)\sin(u)}{1+\cos^2(u)}du$

quartz moonBOT
high zenith
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@steep spire

manic lintel
manic lintel
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pi²/4 was my conclusion too

steep spire
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$2I=\pi\int_0^{\pi}\frac{\pi\sin(u)}{1+\cos^2(u)}du\overset{v=\cos(u)}{=}\pi\int_{-1}^1\frac{1}{1+v^2}dv=\pi\arctan(v)\bigg\vert_{-1}^1=\pi\cdot\frac{\pi}2=\frac{\pi^2}2\implies \boxed{I=\frac{\pi^2}{4}}$.

quartz moonBOT
steep spire
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this is how i did it

high zenith
high zenith
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U middle school

steep spire
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yes

high zenith
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Which country

steep spire
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guess

high zenith
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China

steep spire
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keep on guessin

high zenith
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It's America

steep spire
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nope

high zenith
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India

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Korea

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Japan

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Singapore

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Mars

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Andromeda

high zenith
steep spire
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yeah andromeda

high zenith
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Bro what age ur

high zenith
steep spire
steep spire
high zenith
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And not in middle school but high school

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And u don't know Spanish

steep spire
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2 truths and a lie, which one do you choose?

high zenith
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U learn Spanish on duolingo

steep spire
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good boy

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so which's the lie?

high zenith
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U r 18

steep spire
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nope

high zenith
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Chose one finger ✌️

high zenith
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Jus two question
R u preparing for jee adv thru coaching institute
If yes then how long has it been

zealous tangleBOT
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@high zenith

<:HelpIcon:1304095958283321385>| Help Reminder

Hello raunak._.k, this is a friendly reminder that your help request has been inactive for more than 24 hours. If you no longer need assistance, please consider closing the thread using the +close command. This thread will be automatically closed in 3 days if it remains inactive.