The three basic trigonometric substitutions are in the table below. Try using Substitution I= a sin 0, Integral Contains va Va +r? I=a tan 0. H=a sec 0, 0

The three basic trigonometric substitutions are in the table below.<br>Try using Substitution<br>I= a sin 0,<br>Integral Contains<br>va<br>Va +r?<br>I=a tan 0.<br>H=a sec 0,<br>0<0< * or<br>Part 1.<br>Using the substitution: I = tan 0,- <0 <<br>re-write the indefinite integral then evaluate in terms of 0.<br>dr =<br>V1+z?<br>Note: type 'theta' for 0 and 'dtheta' for do.<br>Part 2.<br>Back substituting in the antiderivative you found in Part 1. above we have<br>dr =<br>V1+z?<br>Note: answer should be in terms of z only.<br>

Extracted text: The three basic trigonometric substitutions are in the table below. Try using Substitution I= a sin 0, Integral Contains va Va +r? I=a tan 0. H=a sec 0, 0<>< *="" or="" part="" 1.="" using="" the="" substitution:="" i="tan" 0,-=""><0>< re-write="" the="" indefinite="" integral="" then="" evaluate="" in="" terms="" of="" 0.="" dr="V1+z?" note:="" type="" 'theta'="" for="" 0="" and="" 'dtheta'="" for="" do.="" part="" 2.="" back="" substituting="" in="" the="" antiderivative="" you="" found="" in="" part="" 1.="" above="" we="" have="" dr="V1+z?" note:="" answer="" should="" be="" in="" terms="" of="" z="">

Jun 08, 2022

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The three basic trigonometric substitutions are in the table below. Try using Substitution I= a sin 0, Integral Contains va Va +r? I=a tan 0. H=a sec 0, 0

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