Integratin ∫ tan 3 ⁡ ( x ) d x in two different ways gives two different answersI was trying to find the antiderivative of a function ∫ tan 3 ⁡ ( x ) d x.However, due to substitution differences, my book has a answer of 1 2 tan 2 ⁡ ( x ) + ln ⁡ ( cos ⁡ x ) + C while I got an answer 1 2 sec 2 ⁡ ( x ) + ln ⁡ ( cos ⁡ x ) + C.The problem is what to substitute in ∫ tan ⁡ x sec 2 ⁡ x d x. The book puts tan ⁡ x = z, while I put sec ⁡ x = z. I don't know if both are correct. If they are, can all functions have multiple antiderivatives?

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Integratin   ∫      tan    3    ⁡  (  x  )    d  x in two different ways gives two different answersI was trying to find the antiderivative of a function   ∫      tan    3    ⁡  (  x  )    d  x.However, due to substitution differences, my book has a answer of       1    2        tan    2    ⁡  (  x  )  +  ln  ⁡  (  cos  ⁡  x  )  +  C while I got an answer       1    2        sec    2    ⁡  (  x  )  +  ln  ⁡  (  cos  ⁡  x  )  +  C.The problem is what to substitute in   ∫  tan  ⁡  x      sec    2    ⁡  x  d  x. The book puts   tan  ⁡  x  =  z, while I put   sec  ⁡  x  =  z. I don&#039;t know if both are correct. If they are, can all functions have multiple antiderivatives?

sepolturamo · Accepted Answer

Step 1The functions       sec    2    ⁡      x   and       tan    2    ⁡      x   differ by a constant. This means that when you have       sec    2    ⁡      x   plus a constant, it can always be turned into an expression containing       tan    2    ⁡      x   plus another constant and vice versa. For example:      sec    2    ⁡      x    +  5  =      (          tan      2        ⁡          x        +    1    )    +  5  =      tan    2    ⁡      x    +  6.More generally:      sec    2    ⁡      x    +      C    1    =      (          tan      2        ⁡          x        +    1    )    +      C    1    =      tan    2    ⁡      x    +      C    2    .In your case:      1    2        tan    2    ⁡      x    +  ln  ⁡      cos    ⁡    x    +  C  =        1    2        (          sec      2        ⁡          x        +    1    )    +  ln  ⁡      cos    ⁡    x    +  C  =        1    2        sec    2    ⁡      x    +      1    2    +  ln  ⁡      cos    ⁡    x    +  C  =        1    2        sec    2    ⁡      x    +  ln  ⁡      cos    ⁡    x    +      (          1      2        +    C    )    =        1    2        sec    2    ⁡      x    +  ln  ⁡      cos    ⁡    x    +      C    1    .Since       C    1   is just another constant, they still use C to denote that.

Denisse Valdez · Accepted Answer

Step 1When evaluating indefinite integrals, there&#039;s no such thing as an incorrect substitution, just one that won&#039;t help much. Sometimes, more than one helpful option exists. If we try evaluating   ∫  tan  ⁡  x      sec    2    ⁡  x  d  x with   z  =  tan  ⁡  x, we get   ∫  z  d  z  =      1    2        z    2    +  C  =      1    2        tan    2    ⁡  x  +  C, so you might call this a useful substitution. The substitution   z  =  sec  ⁡  x proves equally helpful (but your textbook was hardly going to solve the problem twice!), with the integral becoming   ∫  z  d  z  =      1    2        z    2    +  K  =      1    2        sec    2    ⁡  x  +  K. Step 2If you equate the two antiderivatives from these approaches, you of course get   C  =      1    2    +  K from       sec    2    ⁡  x  =      tan    2    ⁡  x  +  1, so the results don&#039;t disagree at all. (Comparing results for       tan    3    ⁡  x also requires you to use this identity.)

Integratin ∫ tan 3 </msup> ⁡ ( x

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