1.∫[x−x22][x+x22]dx=∫[x2−x44]dx=3x3+3x34+c2.∫e5x[7e2x+e3x3]dx=∫[7e7x+3e2x]dx=49e7x+23e2x+c3.∫(4−3x)31dxput4−3x=t=>23x−3dx=dt⟹dx=3−2(4−t)dt∫(4−3x)31dx=3−2∫t34−tdt=3−2∫(4t−3−t−2)dt=3−2(t2−2+3t31)+c=3−2((4−3x)2−2+3(4−3x)31)+c4.∫0π/4(tanx)3(secx)3dxput,secx=t=>tanxsecxdx=dtand using tan2x=1−sec2x,we get∫0π/4(tanx)3(secx)3dx=∫0π/4(secx)2(tanx)2tanxsecxdx=∫0π/4(secx)2(1−sec2x)tanxsecxdx=∫12t2(1−t2)dt=∫12(t2−t4)dt=[3t3−5t5]12=−0.32
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