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

hello njini, plz ask 1 ques in a thread.

so i ll ans your 1^st 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