Integration of $sin^2(x)$ using trig substitution

Question

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.

Answer 1

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.

Answer 2

Putting $\tan x=u,$

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

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

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

Answer 3

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.

Answer 4

$$ \int\sin^{2}\left(x\right)\,{\rm d}x = \int{1 - \cos\left(2x\right) \over 2}\,{\rm d}x = {1 \over 2}\,x - {1 \over 4}\,\sin\left(2x\right) + \mbox{constant} $$