Click to Chat

1800-1023-196

+91-120-4616500

CART 0

• 0

MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping
`        Past exam paper. Which needs answers.            Help is appreciated very much. Thank you all in advannce`
one month ago

1669 Points
```							hello njini, plz ask 1 ques in a thread.so i ll ans your 1st ques.let J= ∫dp/(3 – 4sinp + 2cosp) from 0 to pithis can be easily solved by substituting t= tan(p/2) so that 2dt/dp= sec^2(p/2)= 1+ tan^2(p/2)= 1+t^2 and using the standard formulas:sinp= 2t/(1+t^2) and cosp= (1 – t^2)/(1+t^2)so putting in all these values you shall obtain a quad expression in denominator which can be factorised and converted to partial fractions to obtain the final ans as: – 0.71484 approxkindly approve :=)
```
one month ago
Think You Can Provide A Better Answer ?

## Other Related Questions on Algebra

View all Questions »

### Course Features

• 731 Video Lectures
• Revision Notes
• Previous Year Papers
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Test paper with Video Solution

### Course Features

• 101 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions