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  • Confused! kindly explain, - Integral Calculus - JEE Main-3

What is the reduction formula for \int_{0}^{\frac{\pi }{4}}\tan ^{n}xdx ?

  • Option 1)

    \left ( n-1 \right )\left ( I_{n}+I_{n-1} \right )=1

  • Option 2)

    \left ( n-1 \right )\left ( I_{n}+I_{n-2} \right )=1

  • Option 3)

    n\left ( I_{n}+I_{n-1} \right )=1

  • Option 4)

    n\left ( I_{n}+I_{n-2} \right )=1

 

Answers (1)

best_answer

As we learned,

 

Reduction formulae -

\int sin^{n}xdx=I_{n}   (let)

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

 

-

 

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

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

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

I_{n}=\frac{\left [ \left [ \tan x \right ]^{n-1} \right ]^{\frac{\pi }{2}}_{0}}{n-1}-I_{n-2}

Thus  I_{n}=\frac{1}{n-1}-I_{n-2}\Rightarrow \left ( n-1 \right )\left ( I_{n}+I_{n-2} \right )=1


Option 1)

\left ( n-1 \right )\left ( I_{n}+I_{n-1} \right )=1

Option 2)

\left ( n-1 \right )\left ( I_{n}+I_{n-2} \right )=1

Option 3)

n\left ( I_{n}+I_{n-1} \right )=1

Option 4)

n\left ( I_{n}+I_{n-2} \right )=1

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