∫dx/rootover(sin^3x.cos^5x)= 1.2/3((tan^2x-1)/roo

Question

∫dx/rootover(sin^3x.cos^5x)=1.2/3((tan^2x-1)/root tanx)+c2.2/3((tan^2x-2)/root tanx)+c3.2/3((tan^2x-3)/root tanx)+c4.2/3((tan^2x-4)/root tanx)+cplease solve this the answer is option 3please explain me the procedure

Arun · Accepted Answer

Integral of 1/sqrt of (sin^3xcos^5x) dx=integral of sec^4x dx/sqrt of tan^3xlet tanx=ton differentiating w.r.to x we getsec^2x dx=dtintegral 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

Integral Calculus> ∫dx/rootover(sin^3x.cos^5x)= 1.2/3((tan^2...

∫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

chandrika , 6 Years ago

Arun

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Integral Calculus> ∫dx/rootover(sin^3x.cos^5x)= 1.2/3((tan^2...

∫dx/rootover(sin^3x.cos^5x)=1.2/3((tan^2x-1)/root tanx)+c2.2/3((tan^2x-2)/root tanx)+c3.2/3((tan^2x-3)/root tanx)+c4.2/3((tan^2x-4)/root tanx)+cplease solve this the answer is option 3please explain me the procedure

chandrika , 6 Years ago

Arun

LIVE ONLINE CLASSES

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The area of a square field is 640000cm2.what is its area in hectares

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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
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please explain me the procedure

The area of a square field is 640000cm².what is its area in hectares

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