Consider this integral $\displaystyle I=\int { \tan { x } \sec ^{ 2 }{ x } } dx$.

I solved it in this way $let,\quad \sec { x } =s\\ then,\quad \tan { x } \sec { x } dx=ds\\ \therefore I=\int { s } ds=\cfrac { { s }^{ 2 } }{ 2 } +c=\cfrac { \sec ^{ 2 }{ x } }{ 2 } +c$

is it wrong?

Because my book had this solution: $I=\int { \tan { x } \sec ^{ 2 }{ x } } dx\\ let,\quad \tan { x } =t\\ then,\quad \sec ^{ 2 }{ x } dx=dt\\ I=\int { t } dt=\cfrac { { t }^{ 2 } }{ 2 } +c=\cfrac { \tan ^{ 2 }{ x } }{ 2 } +c$

How come?

Update!

Ok, I got it thanx Brilliant community.

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## Comments

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TopNewestBoth of the approaches here are absolutely correct. It's just that the $c$'s are different.

If the constants of integration in the first approach and the second approach are $c_1$ and $c_2$ respectively, we have $c_1+\frac{1}{2}=c_2$.

Does this help?

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Thanx for replying.

So, beacuse of the constant will the definite integral be different?

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Uh no. Because the difference in the constants get cancelled out on a definite integral. :-0 o_O

So if it was something like $\int _{ a }^{ b }{ tanx{ sec }^{ 2 }x\quad dx } = \frac{ {sec}^{2}b}{2 } + { c }_{ 1 } - \frac{ {sec}^{2}a}{2 } - { c }_{ 1 }$ and similar thing will happen with $\frac{{tan}^2{x}}{2} + {c}_{2}$.

And because the difference between $\frac{{tan}^{2}x}{2}$ and $\frac{{sec}^{2}x}{2}$ is a constant. The answer will be the same.

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Yo! :P If you're curious, feel free to check out Trevor's note on this topic! :D

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thanx for replying. :)

i will look into the note.

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Both answers absolutely correct since, adding 1/2 to second answer (TANx^2) and subtracting 1/2 from c won't matter...

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both answers are ultimately the same. from the infinite set of c if you take out 0.5 and add it to the answer given in the book, it leads to the answer you found out! As simple as that!

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What is equal to i^2n+2

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If 'i' here is considered as the imaginary number then the value depends strictly on n. For n $in$ Z (the set of integers), the value will alternate between -1 and 1.

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The solution given in the book is right

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the c in the book method will be 1/2 less than the c in your method

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i dont understand the formula

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