Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa

Evaluate  \int_{0}^{\pi} \sin^{3}t\cos^{3}t \;dt.

  • Option 1)

    \frac{\pi}{48}

  • Option 2)

    \frac{\pi}{24}

  • Option 3)

    0

  • Option 4)

    None of these

 

Answers (1)

best_answer

As we have learnt,

 

Walli's Method -

Definite integral by first principle

\int_{a}^{b}f(x)dx= \left ( b-a \right )\lim_{n \to \infty }\frac{1}{n}\left [ f(a) +f(a+h)+f(a+2h)....\right ]

where

h=\frac{b-a}{n}

- wherein

 

 \int_{0}^{\pi} \sin^{3}t\cos^{3}t\;dt = \int_{0}^{\pi} (\sin t\cos t)^{3}\;dt = \frac{1}{8}\int_{0}^{\pi}(\sin 2t)^{3}\; dt = 0

 


Option 1)

\frac{\pi}{48}

Option 2)

\frac{\pi}{24}

Option 3)

0

Option 4)

None of these

Posted by

prateek

View full answer

JEE Main high-scoring chapters and topics

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

Similar Questions

Latest Question