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I was solving the integral $$\int\frac{\sin x}{\cos^3x}dx$$ I used the substitution $$u=\cos x \qquad du=-\sin x\,dx \qquad -du=\sin x dx$$ So the integral has the form $$\int-\frac{du}{u^3}=\frac{1}{2u^2}+C=\frac{1}{2\cos^2x}+C$$ However, when I checked the answer in the book, this was $$\int\frac{\sin x}{\cos^3x}dx=\frac{\tan^2x}{2}+C$$ Because it was solved using the conversion $$\int\frac{\sin x}{\cos^3x}=\int \tan x\sec^2x$$ And the substitution $u=\tan x$. So, I derived both functions and the result is the same.

Is this normal? Does this mean that both equations are equal? Can I use any of these answers?

Thank you for reading!

Blue
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Juancho Arias
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    Those two answers differ by a constant. – jjagmath Oct 13 '24 at 02:01
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    What @jjagmath said is true. Think about $\tan^2{x}+1=\sec^2{x}$ and it all makes sense. – Red Five Oct 13 '24 at 02:03
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    At some level these integrals serve as a proof they differ by a constant but if you're new to integration and don't trust it then it might be reasonable for you to try to find an alternative proof that those two functions differ by a constant. To get some intuition (not a proof) start by graphing both of them after fixing constants. – Sidharth Ghoshal Oct 13 '24 at 02:05

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There's nothing wrong with what you did. Your answer is just the same as what is in the answer key (they only differ by a constant).

To prove this, let's show that: $$\frac{1}{2\cos^{2}{x}}+C_1=\frac{\tan^{2}{x}}{2}+C_2$$

Let's focus on the left side: $$\frac{1}{2\cos^{2}{x}}+C_1=\frac{1}{2}\cdot\sec^2{x}+C_1=\frac{1}{2}\cdot\left(1+\tan^{2}{x}\right)+C_1=\frac{\tan^{2}{x}}{2}+C_{1}+\frac{1}{2}$$ Since the sum of constants is just a constant, we can let $C_1+\frac{1}{2}=C_{2}$. Therefore, $$\frac{1}{2\cos^{2}{x}}+C_1=\frac{\tan^{2}{x}}{2}+C_{2}$$

Hanlee
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