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Past exam paper. Which needs answers.            Help is appreciated very much. Thank you all in advannce
5 months ago

1815 Points

so i ll ans your 1st ques.
let J= ∫dp/(3 – 4sinp + 2cosp) from 0 to pi
this 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 approx
kindly approve :=)
5 months ago
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• 101 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions