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I tried to find a generic antiderivative for

$$\displaystyle \int \sec^2x \tan x \mathop{dx} $$

but I think there is something wrong with my solution because it doesn't match what I got through an online calculator.

What am I doing wrong?

Below is my solution.

We will use substitution:

$$u = \sec x \qquad du = \sec x \tan x \, dx$$

We substitute and apply the power rule:

$$ \int (\sec x) (\sec x \tan x \, dx) = \int u \, du = \frac{1}{2} u^2 + C = \frac{\sec^2x}{2} + C$$

The solution I found with the online calculator is:

$$ \frac{\tan ^2 x}{2} + C$$

The steps in the online solution make sense also, so I'm not sure what's going on.

The one thing I have some doubts about is whether I derived the $du$ from $u = \sec x$ correctly. But it seems okay to me. I used implicit differentiation with $x$.

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    $\begingroup$ $\sec^2{(x)}=1+\tan^2{(x)}=\tan^2{(x)}+C$ hence these functions differ by a constant... $\endgroup$ – Peter Foreman Jul 23 at 18:23
  • $\begingroup$ What happens if you try $u=\tan \theta?$ $\endgroup$ – Chris Leary Jul 23 at 18:25
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$$\frac{\sec^{2}(x)}{2} + C = \frac{1+\tan^{2}(x)}{2} + C = \frac{1}{2} + \frac{\tan^{2}(x)}{2} + C$$

$\frac{1}{2}$ is just another constant, in indefinite integral constant doesn't really matter unless you're asked for the integrand original function so you can just kind of "combine" $\frac{1}{2}$ into C, you can also differentiate the answer to know that there's nothing wrong at all with your answer

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Your answer is correct.

The difference between your answer and the answer given is that they differ by a constant value.

You can see this by using the identity

$$1+\tan^2(x) = \sec^2(x)$$

Hence your answer can be converted to the given answer by subtracting $-1/2$, which is a constant.

As mentioned in the comments and in another answer you can also directly get the form in the answer by using the substitution

$$u = \tan x$$

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  • $\begingroup$ I modified your answer a bit to use MathJax. Going forward, please use MathJax for mathematical typesetting for ease of readability. Good answer though! $\endgroup$ – Cameron Williams Jul 23 at 18:29
  • $\begingroup$ Thanks for editing the expressions $\endgroup$ – StackUpPhysics Jul 23 at 18:29
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Try differentiating both of them and see that nothing has gone wrong at all.

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Let $u = \tan x$.

Then $du = \sec^{2} x dx$.

Hence, the integral becomes

$$ \int u du = \frac{u^{2}}{2} + C = \frac{\tan^{2}x}{2} + C. $$

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