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.

No vote yet

1 vote

×

Problem Loading...

Note Loading...

Set Loading...

Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## 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?

Log in to reply

Thanx for replying.

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

Log in to reply

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.

Log in to reply

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

Log in to reply

thanx for replying. :)

i will look into the note.

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!

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...

Log in to reply

i dont understand the formula

Log in to reply

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

Log in to reply

The solution given in the book is right

Log in to reply

What is equal to i^2n+2

Log in to reply

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