OJEE King's Rule — practice questions

Explore 2 free King's Rule questions in OJEE Mathematics. Open each question or exercise to read its content, reference answers and available explanation.

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King's Rule questions and exercises with answers

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$\displaystyle\int_0^{\pi/2} \dfrac{\sin x}{\sin x + \cos x}\, dx = ?$Read the answer and worked solution →$\displaystyle\int_0^\pi \dfrac{x\sin x}{1 + \cos^2 x}\, dx = ?$Read the answer and worked solution →

Worked practice questions

$\displaystyle\int_0^{\pi/2} \dfrac{\sin x}{\sin x + \cos x}\, dx = ?$

$0$
$\pi/4$
$\pi/2$
$1$

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Correct answer: B. $\pi/4$

King's rule (substitute $x \to \pi/2 - x$): $I = \int_0^{\pi/2} \cos x/(\cos x + \sin x)\, dx$. **Add to original:** $2I = \int_0^{\pi/2} (\sin x + \cos x)/(\sin x + \cos x)\, dx = \pi/2$. **So $I = \pi/4$.**

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$\displaystyle\int_0^\pi \dfrac{x\sin x}{1 + \cos^2 x}\, dx = ?$

$\pi^2/4$
$\pi/4$
$\pi^2/8$
$\pi$

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Correct answer: A. $\pi^2/4$

Use $\int_0^\pi x f(\sin x)\, dx = (\pi/2)\int_0^\pi f(\sin x)\, dx$. Here $f(\sin x) = \sin x/(1 + \cos^2 x)$. Now sub $u = \cos x$, $du = -\sin x\, dx$: $\int_0^\pi \sin x/(1+\cos^2 x)\, dx = \int_{-1}^{1} du/(1+u^2) = [\arctan u]_{-1}^1 = \pi/2$. **Final:** $(\pi/2)(\pi/2) = \pi^2/4$.