Help me to understand this integration.

Consider this integral I=tanxsec2xdx\displaystyle I=\int { \tan { x } \sec ^{ 2 }{ x } } dx.

I solved it in this way let,secx=sthen,tanxsecxdx=dsI=sds=s22+c=sec2x2+clet,\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=tanxsec2xdxlet,tanx=tthen,sec2xdx=dtI=tdt=t22+c=tan2x2+cI=\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
4 years, 9 months ago

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1 vote

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Both of the approaches here are absolutely correct. It's just that the cc's are different.

If the constants of integration in the first approach and the second approach are c1c_1 and c2c_2 respectively, we have c1+12=c2c_1+\frac{1}{2}=c_2.

Does this help?

Mursalin Habib - 4 years, 9 months ago

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Thanx for replying.

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

Soumo Mukherjee - 4 years, 9 months ago

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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 abtanxsec2xdx=sec2b2+c1sec2a2c1\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 tan2x2+c2\frac{{tan}^2{x}}{2} + {c}_{2}.

And because the difference between tan2x2\frac{{tan}^{2}x}{2} and sec2x2\frac{{sec}^{2}x}{2} is a constant. The answer will be the same.

Vishnuram Leonardodavinci - 4 years, 9 months ago

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

Vishnuram Leonardodavinci - 4 years, 9 months ago

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thanx for replying. :)

i will look into the note.

Soumo Mukherjee - 4 years, 9 months ago

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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...

ashay wakode - 4 years, 9 months ago

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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!

Wrik Bhadra - 4 years, 9 months ago

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What is equal to i^2n+2

manthar Ali - 4 years, 9 months ago

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

Wrik Bhadra - 4 years, 9 months ago

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The solution given in the book is right

Avishkar Raut - 4 years, 9 months ago

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

Rutwik Dhongde - 4 years, 9 months ago

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i dont understand the formula

shweta dalal - 4 years, 9 months ago

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