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Prove that 0π4(xxsinx+cosx)2dx=4π4+π.\large\int_{0}^{\frac{\pi}{4}} \left(\dfrac{x}{x\sin x+\cos x}\right)^2 dx=\dfrac{4-\pi}{4+\pi}.

#Calculus #DefiniteIntegral

Note by Adarsh Kumar
5 years, 6 months ago

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Comments

By Differentiation - Quotient Rule , the integrand might be in the form of vuuvv2 \dfrac {v u' - u v'}{v^2} , so we can make the assumption that v=xsin(x)+cos(x)v = x\sin(x) + \cos(x) .

Now we just want to find the unknown coefficients for u=Axcos(x)+Bxsin(x)+Csin(x)+Dcos(x)+Exu = A x\cos(x) + Bx \sin(x) + C \sin(x)+ D \cos(x) + Ex such that vuuv=x2vu' - uv' = x^2 . Solving this gives u=sin(x)xcos(x)u = \sin(x) - x\cos(x) .

Checking back, the indefinite integral of x2xsin(x)+cos(x) \dfrac{x^2}{x\sin(x) + \cos(x)} is indeed sin(x)xcos(x)xsin(x)+cos(x) \dfrac{\sin(x) - x\cos(x)}{x\sin(x) + \cos(x)} . Substituting the limits give the desired answer.

Pi Han Goh - 5 years, 6 months ago

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Challenge student note: nice solution. U have used the trick perfectly.

Aditya Kumar - 5 years, 6 months ago

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!

Adarsh Kumar - 5 years, 6 months ago

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Please don't use square brackets if it is not a gif. It confuses.

Aditya Kumar - 5 years, 6 months ago

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@Aditya Kumar Ooops!Sorry!

Adarsh Kumar - 5 years, 6 months ago

I will post this one later. Got it.

Mardokay Mosazghi - 5 years, 6 months ago
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