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.

## Comments

Sort by:

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? – Mursalin Habib · 2 years, 9 months ago

Log in to reply

So, beacuse of the constant will the definite integral be different? – Soumo Mukherjee · 2 years, 9 months ago

Log in to reply

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. – Vishnuram Leonardodavinci · 2 years, 9 months ago

Log in to reply

Yo! :P If you're curious, feel free to check out Trevor's note on this topic! :D – Vishnuram Leonardodavinci · 2 years, 9 months ago

Log in to reply

i will look into the note. – Soumo Mukherjee · 2 years, 9 months ago

Log in to reply

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! – Wrik Bhadra · 2 years, 9 months ago

Log in to reply

Both answers absolutely correct since, adding 1/2 to second answer (TANx^2) and subtracting 1/2 from c won't matter... – Ashay Wakode · 2 years, 9 months ago

Log in to reply

i dont understand the formula – Shweta Dalal · 2 years, 8 months ago

Log in to reply

the c in the book method will be 1/2 less than the c in your method – Rutwik Dhongde · 2 years, 8 months ago

Log in to reply

The solution given in the book is right – Avishkar Raut · 2 years, 9 months ago

Log in to reply

What is equal to i^2n+2 – Manthar Ali · 2 years, 9 months ago

Log in to reply

– Wrik Bhadra · 2 years, 8 months ago

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.Log in to reply