Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • 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

Posted by

Aadil

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JEE Main 2027: NTA allows fresh application forms, warns against multiple IDs; correction window from Nov 27
anam-khan

JEE Main 2027: NTA advises Aadhaar re-KYC on DigiLocker before registration, issues guidelines
anam-khan

Kota: JEE aspirant dies by suicide in PG room; another student's death triggers allegations of negligence

Latest Question