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Definite Integration question

2002 · Shift 0 · Q79

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JEE Main
Mathematics
Definite Integration
2002 · Shift 0 · Q79

Definite Integration question

2002 · Shift 0 · Q79

JEE MainMathematicsDefinite IntegrationMCQ+4 / −1

{I_n} = \int\limits_0^{\pi /4} {{{\tan }^n}x\,dx}

then

\,\mathop {\lim }\limits_{n \to \infty } \,n\left[ {{I_n} + {I_{n + 2}}} \right]

equals

A
${1 \over 2}$
B
$1$
C
$\infty$
D
zero

View written solutionFree

Correct answer: B

We are given $I_n=\int_0^{\pi/4} \tan^n x\,dx$ and we need to find $\lim_{n\to\infty} n\,[I_n+I_{n+2}].$
Rewrite the integral using the substitution $t=\tan x,\qquad dx=\frac{dt}{1+t^2}.$ When $x=0$ , $t=0$ ; when $x=\pi/4$ , $t=1$ . So $I_n=\int_0^1 \frac{t^n}{1+t^2}\,dt.$ Similarly, $I_{n+2}=\int_0^1 \frac{t^{n+2}}{1+t^2}\,dt.$
Add them:

=\int_0^1 t^n\,dt,$$ because $$\frac{t^n+t^{n+2}}{1+t^2}=t^n\frac{1+t^2}{1+t^2}=t^n.$$ 4. Now evaluate: $$I_n+I_{n+2}=\int_0^1 t^n\,dt=\frac{1}{n+1}.$$ Therefore, $$n[I_n+I_{n+2}]=\frac{n}{n+1}.$$ 5. Take the limit: $$\lim_{n\to\infty} \frac{n}{n+1}=1.$$ 6. Hence the correct option is $$\boxed{1}.$$ So option **B** is correct.

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