# Help me to understand this integration.

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.

Note by Soumo Mukherjee
3 years, 11 months ago

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Both 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?

- 3 years, 11 months ago

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

- 3 years, 11 months ago

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.

- 3 years, 11 months ago

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

- 3 years, 11 months ago

thanx for replying. :)

i will look into the note.

- 3 years, 11 months ago

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!

- 3 years, 11 months ago

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

- 3 years, 11 months ago

i dont understand the formula

- 3 years, 10 months ago

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

- 3 years, 11 months ago

The solution given in the book is right

- 3 years, 11 months ago

What is equal to i^2n+2

- 3 years, 11 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.

- 3 years, 10 months ago