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Hi I'm just learning integration, so I'm sorry if this is really basic.

I know this is true: $\displaystyle \int sin^2(x)d x = \frac{1}{2} (x - sin(x)cos(x)) + C$

Because you can use Power reducing substitution http://www.youtube.com/watch?v=VRNuPqA_Vo8

But my question is why can't you use trig substitution like below?

$$\int sin^2x$$ $$= \int \frac{tan^2x}{sec^2x} $$ $$Let\;a = tan x$$ $$As sin^2(x)+cos^2(x)=1...$$ $$= \int \frac{a^2}{1+a^2}$$ $$= 1-tan^-1(a)$$ $$= 1-x$$

Please let me know where I'm assuming something incorrect.

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    $\begingroup$ Interesting idea! If you let $a=\tan x$, then $da=\sec^2 x\,dx$, so $dx=\frac{da}{\sec^2 x}=\frac{da}{1+a^2}$. We get a somewhat unattractive integral $\int \frac{a^2}{(1+a^2)^2}\,da$. By the way, you should not leave out $dx$ in your notation for integrals. Leaving it out is (i) wrong and (ii) is likely to lead to error in substitutions. $\endgroup$ – André Nicolas Oct 12 '13 at 18:23
  • $\begingroup$ @AndréNicolas Thanks, I'll try to remember that, always leave out the +c also! $\endgroup$ – RSenApps Oct 12 '13 at 18:23
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Putting $\tan x=u,$

$\displaystyle x=\arctan u\implies dx=\frac{du}{1+u^2}$

and $\sec^2x=1+\tan^2x=1+u^2$

$$\implies\int\sin^2xdx=\int\frac{\tan^2x}{\sec^2x}dx=\int\frac{u^2}{(1+u^2)^2}du$$

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    $\begingroup$ Thanks, I know the proper way to do this, I was just wondering where my logic is flawed? $\endgroup$ – RSenApps Oct 12 '13 at 18:12
  • $\begingroup$ ah ok that makes sense, was forgetting about dx... Thanks for your help it says I need to wait 3 minutes to accept your answer $\endgroup$ – RSenApps Oct 12 '13 at 18:19
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    $\begingroup$ @RSenApps, sorry for misinterpreting the original question $\endgroup$ – lab bhattacharjee Oct 12 '13 at 18:20
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It looks like you're trying to assert that

$$\sin^2(x) = \tan^2(x)\sec^2(x)$$

in the first step. But,

$$\tan^2(x)\sec^2(x) = \frac{\sin^2(x)}{\cos^2(x)}\frac{1}{\cos^2(x)} = \frac{\sin^2(x)}{\cos^4(x)} \neq \sin^2(x).$$

EDIT: Ok, now that it is properly formatted, you are correctly asserting that

$$ \sin^2(x) = \frac{\tan^2(x)}{\sec^2(x)}. $$

However, when you do the substitution $a = \tan(x)$ you need $da = \sec^2(x)dx$. You have $\sec^2(x)$ in the denominator, rather than the numerator, so you cannot do this substitution.

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    $\begingroup$ Oh, you just changed the question! Let me check again! $\endgroup$ – Jeremy West Oct 12 '13 at 18:16
  • $\begingroup$ ah ok that makes sense, was forgetting about dx... $\endgroup$ – RSenApps Oct 12 '13 at 18:20
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The integral you are trying to solve is $\int tan^2(x) dx$. After your substitution you get $\int \frac{a^2}{1+a^2} dx$. You have to express $dx$ in terms of $da$ if you want to make any sense of this.

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  • $\begingroup$ ah ok that makes sense, was forgetting about dx... $\endgroup$ – RSenApps Oct 12 '13 at 18:20
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$$ \int\sin^{2}\left(x\right)\,{\rm d}x = \int{1 - \cos\left(2x\right) \over 2}\,{\rm d}x = {1 \over 2}\,x - {1 \over 4}\,\sin\left(2x\right) + \mbox{constant} $$

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