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  • Stuck here, help me understand: - Integral Calculus - JEE Main-2

I_{n}=\int_{0}^{\pi /4}tan^{n}x\, dx,\; then\; \lim_{n\rightarrow \infty }n\left [ I_{n}+I_{n-2} \right ] \; equals

  • Option 1)

    1/2

  • Option 2)

    1

  • Option 3)

    \infty

  • Option 4)

    0

 

Answers (1)

best_answer

As learnt in concept

Reduction formulae -

\int sin^{n}xdx=I_{n}   (let)

Then nI_{n}-(n-1)I_{n-2}=- \sin^{n-1}x\cos x

 

-

 

 We will use the reduction formulas

I_{n}=\int_{0}^{\frac{\pi}{4}}tan^{n}x\ dx

\lim_{n \to \infty }[I_{n}+I_{n-2}]=?

I_{n}+I_{n-2}=\int_{0}^{\frac{\pi}{4}}tan^{n-2}x(1+tan^{2})dx

=\int_{0}^{\frac{\pi}{4}}tan^{n-2}x\ sec^{2}xdx

=\left [\frac{tan^{n-1}x}{n-1} \right ]^{\frac{\pi}{4}}_{0}=\frac{1}{n-1}

\lim_{n \to \infty}n[I_{n}+I_{n-2}]=\frac{n}{n-1}

=\lim_{n \to \infty}\frac{1}{1-\frac{1}{n}}=1


Option 1)

1/2

This is incorrect

Option 2)

1

This is correct

Option 3)

\infty

This is incorrect

Option 4)

0

This is incorrect

Posted by

prateek

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