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

Now we just want to find the unknown coefficients for \(u = A x\cos(x) + Bx \sin(x) + C \sin(x)+ D \cos(x) + Ex \) such that \(vu' - uv' = x^2 \). Solving this gives \(u = \sin(x) - x\cos(x) \).

Checking back, the indefinite integral of \( \dfrac{x^2}{x\sin(x) + \cos(x)} \) is indeed \( \dfrac{\sin(x) - x\cos(x)}{x\sin(x) + \cos(x)} \). Substituting the limits give the desired answer.
–
Pi Han Goh
·
1 year, 1 month ago

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@Pi Han Goh
–
Challenge student note: nice solution. U have used the trick perfectly.
–
Aditya Kumar
·
1 year, 1 month ago

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@Pi Han Goh
–
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
·
1 year, 1 month ago

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@Adarsh Kumar
–
Please don't use square brackets if it is not a gif. It confuses.
–
Aditya Kumar
·
1 year, 1 month ago

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

Now we just want to find the unknown coefficients for \(u = A x\cos(x) + Bx \sin(x) + C \sin(x)+ D \cos(x) + Ex \) such that \(vu' - uv' = x^2 \). Solving this gives \(u = \sin(x) - x\cos(x) \).

Checking back, the indefinite integral of \( \dfrac{x^2}{x\sin(x) + \cos(x)} \) is indeed \( \dfrac{\sin(x) - x\cos(x)}{x\sin(x) + \cos(x)} \). Substituting the limits give the desired answer. – Pi Han Goh · 1 year, 1 month ago

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– Aditya Kumar · 1 year, 1 month ago

Challenge student note: nice solution. U have used the trick perfectly.Log in to reply

– Adarsh Kumar · 1 year, 1 month 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!Log in to reply

– Aditya Kumar · 1 year, 1 month ago

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

– Adarsh Kumar · 1 year, 1 month ago

Ooops!Sorry!Log in to reply

I will post this one later. Got it. – Mardokay Mosazghi · 1 year, 1 month ago

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