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Question Number 88859 by 242242864 last updated on 13/Apr/20
∫eaxcosbxdx∫x2e2xln3x2dx
Commented by john santu last updated on 13/Apr/20
1)I=∫eaxcosbxdxI=1beaxsinbx−1ab∫eaxsinbxdxI=eaxsinbxb−1ab[−eaxcosbxb+1ab∫eaxcosbxdx]I=eaxsinbxb+eaxcosbxab2−1(ab)2∫eaxcosbxdx(1+1(ab)2)I=eax(absinbx+cosbx)ab2I=a2b2eax(absinbx+cosbx)(1+(ab)2)(ab2)I=aeax(absinbx+cosbx)(1+(ab)2)
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