Integral Calculus-2
Integral Calculus-2
Allen Star Batch
Grade 12
Question:
If $P = \int_0^\infty \frac{x^2}{1+x^4} dx$; $Q = \int_0^\infty \frac{xdx}{1+x^4}$ and $R = \int_0^\infty \frac{dx}{1+x^4}$, then:
$Q = \frac{\pi}{4}$
$P = R$
$P - \sqrt{2}Q + R = \frac{\pi}{2\sqrt{2}}$
$P - 2\sqrt{2}Q + R = \frac{\pi}{\sqrt{2}}$
Step-by-Step Solution
Key Concept: Clever substitution and symmetry properties reduce the integral to arctangent form, which evaluates directly.
For $Q = \int_0^\infty \frac{x\,dx}{1+x^4}$, substitute $x^2 = t$ to get $Q = \frac{1}{2}\int_0^\infty \frac{dt}{1+t^2} = \frac{1}{2}\tan^{-1}(t)\Big|_0^\infty = \frac{\pi}{4}$. For $P = \int_0^\infty \frac{x^2}{1+x^4}dx$, substitute $x = 1/t$ to show $P = \int_0^\infty \frac{dt}{1+t^4}$. Adding $P + Q$ and using $1 + \frac{1}{x^2} = \left(x - \frac{1}{x}\right)^2 + 2$ yields $P + Q = \sqrt{2}\cdot\frac{\pi}{2\sqrt{2}} = \frac{\pi\sqrt{2}}{2}$, so $P = \frac{\pi\sqrt{2}}{2} - \frac{\pi}{4}$.
Correct Answer: 1,2,3,4