3
$\begingroup$

$\int \tan^2xdx$

Here is what I tried...

$\int \tan^2xdx = \int (\sec^2x-1)dx = \int(\frac{1}{\cos^2x} -1)dx = \int \frac{1}{ \frac{1}{2} (1+\cos2x)} = \frac{2}{1} \int \frac{1}{1+\cos2x}$

$u = \cos2x, du = 2\sin2xdx$

$= 2 \bigg( \frac{1}{x + \frac{1}{2}\sin2x} \bigg) + C$

But the answer is $ \tan x - x + C$.

I think the way that I set up the problem (using half angle identity of $\cos^2x$ in the denominator started making the calculations a lot harder to follow in my opinion) was an issue and that I could have set it up better?

$\endgroup$
  • 2
    $\begingroup$ I think the main point is that you should know that the integral of the $\sec^2$ is equal to $\tan$. You can derive it probably by setting $u=\cos(x)$ and you get something that simplifies. (You should do the substitution directly without simplifying the $\cos^2$) $\endgroup$ – Stan Tendijck Mar 6 at 14:06
  • 2
    $\begingroup$ There are correct answers below, but you should know that there are things wrong with your answer, that you really should understand. Minor issue (I assume): you lost the $-1$ in the integrand. Major issue (if I understand what you did): $$\int {1\over \text {something} }\,dx \not= {1 \over \int \text{something}\, dx}.$$ $\endgroup$ – peter a g Mar 6 at 14:09
  • $\begingroup$ @peterag yeaaa that's what I did...damn. I can't believe I overlooked derivative of $/tan $ is $\sec^2$ $\endgroup$ – Evan Kim Mar 6 at 14:20
4
$\begingroup$

$$\int(\sec^2x-1)dx=\int\sec^2x\ dx-\int dx=?$$

$\endgroup$
2
$\begingroup$

The derivative of $\tan x$ is $1+\tan^2x$. Then $$\int\tan^2x\,dx = \int\frac{\mathrm{d}}{\mathrm{d}x}\tan x\,\mathrm{d}x-x+c = \tan x -x+c. $$

$\endgroup$
2
$\begingroup$

Use the fact that:

$$ \frac{d}{dx}\left(\tan{x}\right)=\sec^2{x}\implies \int\sec^2{x}\,dx=\tan{x}+C $$

$$ \int \tan^2x\,dx = \int(\sec^2x-1)\,dx= \int\sec^2x\,dx-\int\,dx=\tan{x}-x+C. $$

$\endgroup$
0
$\begingroup$

$$I=\int\tan^2(x)dx=\int\sec^2(x)-1dx$$ as:$$\cos^2(x)+\sin^2(x)=1\to1+\tan^2(x)=\frac{1}{\cos^2(x)}$$ also we know that: $$\frac{d}{dx}\left[\tan(x)\right]=\sec^2(x)$$ so: $$I=\tan(x)-\int dx=\tan(x)-x+C$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.