WBJEE King's Invariance — practice questions

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

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King's Invariance questions and exercises with answers

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$\displaystyle\int_0^{\pi/2} \dfrac{dx}{1 + \tan^3 x} = ?$Read the answer and worked solution →Given $\displaystyle\int_0^{\pi/2} \dfrac{\sin^n x}{\sin^n x + \cos^n x}\, dx$, the value is:Read the answer and worked solution →

Worked practice questions

$\displaystyle\int_0^{\pi/2} \dfrac{dx}{1 + \tan^3 x} = ?$

$\pi/4$
$\pi/2$
$0$
$\ln 2$

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Correct answer: A. $\pi/4$

Same King's trick as $\sin/(\sin+\cos)$: $1/(1 + \tan^3 x) = \cos^3 x/(\cos^3 x + \sin^3 x)$. Sub $x \to \pi/2 - x$ swaps $\sin↔\cos$: $I = \int \sin^3 x/(\sin^3 x + \cos^3 x)$. **Add:** $2I = \int_0^{\pi/2} dx = \pi/2$. So $I = \pi/4$. (Indeed, $\int_0^{\pi/2} dx/(1 + \tan^n x) = \pi/4$ for ANY $n$ — striking invariance.)

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Given $\displaystyle\int_0^{\pi/2} \dfrac{\sin^n x}{\sin^n x + \cos^n x}\, dx$, the value is:

Depends on $n$
$\pi/4$ for all $n$
$\pi/2$ for all $n$
$0$

Show answer and solution

Correct answer: B. $\pi/4$ for all $n$

By King's $x \to \pi/2 - x$, integrand swaps to $\cos^n x/(\cos^n x + \sin^n x)$. **Adding to original:** $2I = \int_0^{\pi/2} 1\, dx = \pi/2$, so $I = \pi/4$ regardless of $n$. Beautiful invariance — a recurring JEE Adv test.