# Definitely a problem!

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

2 years, 5 months ago

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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.

- 2 years, 5 months ago

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

- 2 years, 5 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!

- 2 years, 5 months ago

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

- 2 years, 5 months ago

Ooops!Sorry!

- 2 years, 5 months ago

I will post this one later. Got it.

- 2 years, 5 months ago