Definite Integration
Definite Integration
nta_pyq_2025_apr
Grade 12
Question:
If $I = \displaystyle\int_0^{\pi}\frac{\sin^{3/2}x}{\sin^{3/2}x+\cos^{3/2}x}\,dx$, then $\displaystyle\int_0^{2I}\frac{x\sin x\cos x}{\sin^4 x+\cos^4 x}\,dx$ equals:
$\dfrac{\pi^2}{12}$
$\dfrac{\pi^2}{4}$
$\dfrac{\pi^2}{16}$
$\dfrac{\pi^2}{8}$
Step-by-Step Solution
Key Concept: Use the King's property on $I$ to get $2I = \pi/2$, so $2I = \pi/2$. Then for $J = \int_0^{\pi/2} x\sin x\cos x/(\sin^4 x+\cos^4 x)dx$, apply King's property again and use $\tan^2 x = u$ substitution to evaluate.
By King's property on $[0,\pi]$: $I = \int_0^{\pi}\tfrac{\cos^{3/2}x}{\sin^{3/2}x+\cos^{3/2}x}dx$. Adding: $2I = \pi \Rightarrow I = \pi/4$. So $2I = \pi/2$.
Let $J = \int_0^{\pi/2}\dfrac{x\sin x\cos x}{\sin^4 x+\cos^4 x}dx$. Applying King's: $J = \int_0^{\pi/2}\dfrac{(\pi/2-x)\sin x\cos x}{\sin^4 x+\cos^4 x}dx$.
$$2J = \frac{\pi}{2}\int_0^{\pi/2}\frac{\sin x\cos x}{\sin^4 x+\cos^4 x}dx = \frac{\pi}{4}\int_0^{\infty}\frac{dt}{1+t^2} = \frac{\pi}{4}\cdot\frac{\pi}{2}$$
(where $t = \tan^2 x$). Hence $J = \dfrac{\pi^2}{16}$.
Correct Answer: 3