Apply the reduction formula to this integral and you'll find this outcome: I = (1/48)*[15*lnItan(x)+sec(x)I + 40*sec^5(x) - 70*sec^3(x) + 15*sec(x)tan(x)]+C.
How did you get rid of the sec^2 on the outside of the equation? I understand that the sec^2 was equal to the derivative of U but does that automatically cancel it out of the problem when you are done?
Your comment is five years old and you're probably not in school anymore, but just in case anyone needs clarification: When you solve for dx you get dx = du/sec^2x. When you plug dx in the sec^2 cancels out.
Holaaa estan geniales tus videos me ha ayudado mucho podrias por favor poner subtitulos tambien en español? GRACIAS aunque entiendo lo que haces las matematicas son un idioma universal xD saludos desde Venezuela :)
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The integral = integral of (tanx)^2(1 + (tanx)^2) d(tanx) = (1/3)(tanx)^3 + (1/5)(tanx)^5 + C
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How can i integrate 5 tan^4(x) sec^3(x) dx?
Apply the reduction formula to this integral and you'll find this outcome:
I = (1/48)*[15*lnItan(x)+sec(x)I + 40*sec^5(x) - 70*sec^3(x) + 15*sec(x)tan(x)]+C.
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Can someone help with this T.T
Integral of xtan^8(x^2)sec^4(x^2)dx
genius
How did you get rid of the sec^2 on the outside of the equation? I understand that the sec^2 was equal to the derivative of U but does that automatically cancel it out of the problem when you are done?
Your comment is five years old and you're probably not in school anymore, but just in case anyone needs clarification: When you solve for dx you get dx = du/sec^2x. When you plug dx in the sec^2 cancels out.
Help me (integral of tan^3x*sec^3x)dx
Holaaa estan geniales tus videos me ha ayudado mucho podrias por favor poner subtitulos tambien en español? GRACIAS aunque entiendo lo que haces las matematicas son un idioma universal xD saludos desde Venezuela :)
sorry, i don't have a translator to help me with cc in spanish..
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