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∫dx/rootover(sin^3x.cos^5x)= 1.2/3((tan^2x-1)/root tanx)+c 2.2/3((tan^2x-2)/root tanx)+c 3.2/3((tan^2x-3)/root tanx)+c 4.2/3((tan^2x-4)/root tanx)+c please solve this the answer is option 3 please explain me the procedure

∫dx/rootover(sin^3x.cos^5x)=
1.2/3((tan^2x-1)/root tanx)+c
2.2/3((tan^2x-2)/root tanx)+c
3.2/3((tan^2x-3)/root tanx)+c
4.2/3((tan^2x-4)/root tanx)+c
please solve this the answer is option 3
please explain me the procedure

Grade:12th pass

1 Answers

Arun
25763 Points
2 years ago
Integral of 1/sqrt of (sin^3xcos^5x) dx
=integral of sec^4x dx/sqrt of tan^3x
let tanx=t
on differentiating w.r.to x we get
sec^2x dx=dt
integral of (1+t^2)/t^3/2 .dt
=integral {t^(-3/2) +t^(1/2)} dt
=t^(-1/2)/(-1/2) + t^(3/2)/(3/2)
=-2/sqrt t +2/3 .t^(3/2)
=-2/sqrt of tanx + 2/3(tanx)^3/2

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