Question

integration of (2sin2x-cosx) (6-cos2x-4sinx)1/2 dx

Answer 1

Let I = ∫(2 sin 2x − cos x) dx6−cos2x − 4 sin x=∫(4 sin x . cos x − cos x) dx6 − 1 + sin2x − 4 sin x=∫cos x [4 sin x − 1] dxsin2x−4sin x+5put sin x = t⇒cos x dx = dtNow, I = ∫(4t−1)t2−4t+5 dtLet 4t−1 = A×ddt(t2−4t+5) + B⇒4t−1 = A (2t−4) + B⇒4t−1 = 2At − 4A + B⇒2A = 4 ⇒A = 2and −4A + B = −1⇒−8 + B = −1⇒B = 7so,  4t−1 = 2(t2−4t+5) + 7So, I = 2∫(2t−4)(t2−4t+5) dt + 7∫dt(t2−4t+5)   .....(1)Let I1 = ∫(2t−4)(t2−4t+5) dtPut (t2−4t+5) = v⇒(2t−4) dt = dvSo, I1 = ∫dvv = log ∣∣v∣∣ = log∣∣t2−4t+5∣∣Now, I2 =∫dt(t2−4t+5)=∫dt[t2−4t+4] +1=∫dt(t−2)2 + (1)2=tan−1(t−2)So, I = 2log∣∣t2−4t+5∣∣ + 7tan−1(t−2) + CI = 2 log ∣∣sin2x−4 sin x + 5∣∣ + 7 tan−1[sin x − 2] + C
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