Compute the indefinite integral
∫√cos2xsinxdx
My Attempt:
∫√cos2xsinxdx=∫cos2xsin2x√cos2xsinxdx=∫2cos2x−1(1−cos2x)√2cos2x−1sinxdx
Let cosx=t, so that sinxdx=−dt. This changes the integral to
∫(2t2−1)(t2−1)√2t2−1dt=∫(2t2−2)+1(t2−1)√2t2−1dt=2∫dt√2t2−1+∫dt(t2−1)√2t2−1
How can I solve the integral from here?
Answer
∫√cos2xsinx dx=∫√cos2x−sin2xsinx dx[1]=∫√t4−6t2+1t3+t dt[2]=12∫√u2−6u+1u2+u du[3]=∫(y2−6y+1)2(y−1)(y−3)(y+1)(y2+2t−7) dy[4]=∫[1y−1+1y−3−1y+1−16y2+2y−7] dt=∫[1y−1+1y−3−1y+1−16(y+1)2−8] dt
The rest is yours.
Notes :
[1]Use Weierstrass substitution, tan(x2)=t.
[2]Use substitution u=t2.
[3]Use Euler substitution, y−u=√u2−6u+1⇒y=u2−12u−6.
[4]Use partial fractions decomposition.
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