Evaluate integrate 1/sinx (2+cosx) dx. Share with your friends Share 6 Manbar Singh answered this Let I = ∫dxsin x2 + cos x=∫sin x dxsin2x2 + cos x=∫sin x dx1 - cos2x2 + cos xput cos x = t⇒- sin x dx = dt⇒sin x dx = - dtNow, I = -∫dtt + 21 - t2 = ∫dtt + 2t2 - 1 = ∫dtt-1t+1t+2Let 1t-1t+1t+2 = At-1 + Bt+1 + Ct+2⇒1 = At+1t+2 + Bt-1t+2 + Ct-1t+1⇒1 = At2 + 3t + 2 + Bt2+t-2 + Ct2 - 1⇒1 = A+B+Ct2 + 3A + Bt + 2A-2B-C⇒A + B + C = 0 .........1and 3A + B = 0 ............2and 2A - 2B - C = 1 ........3Solving 1, 2 and 3, we getA = 16; B = -12; C = 13 So,1t-1t+1t+2 = 16×1t-1 - 12×1t+1+13×1t+2Integrating both sides, we get∫dtt-1t+1t+2 = 16∫dtt-1 - 12∫dtt+1 + 13∫dtt+2=16logt-1 - 12logt+1 + 13logt+2 + C=16logcos x - 1 - 12logcos x + 1 + 13logcos x + 2 + C 17 View Full Answer
