# Math Help - Integral confusion

1. ## Integral confusion

I'm trying to find the integral $\int\sec^2{\theta}\tan{\theta}\; {d\theta}$. Here is what I did:

Let $u = \tan{\theta}$. Then $\dfrac{du}{d\theta} = \sec^2\theta \Rightarrow {\color{blue}d{\theta} = \dfrac{du}{\sec^2\theta}}$ so that $\int\sec^2{\theta}\tan{\theta}\;{d\theta} = \int\dfrac{\sec^2{\theta}u}{sec^2\theta}\;{du} = \int{u}\;{du} = \frac{u^2}{2}+C$ $\Rightarrow \int\sec^2{\theta}\tan{\theta}\;{d\theta} = \frac{\tan^2\theta}{2}+C$, which is right.

My first question is, is there something wrong with the highlighted step? I have seen people raising questions about $\dfrac{dy}{dx}$ being treated as if they were numbers.

My second question is to do with when I tried to substitute $u$ for $\sec^2\theta$ and got different results than the above:

Let $u = \sec^2\theta$. Then $\dfrac{du}{d\theta} = 2\sec^2\theta\tan\theta \Rightarrow {\color{blue}d{\theta} = \dfrac{du}{2\sec^2\theta\tan\theta}}$ so that $\int\sec^2{\theta}\tan{\theta}\;{d\theta} = \int\dfrac{u\tan\theta}{2u\tan\theta}\;{du} = \frac{1}{2}\int\frac{u}{u}\;{du} = \frac{u}{2}+C$ $\; \Rightarrow \int\sec^2{\theta}\tan{\theta}\;{d\theta} = \frac{\sec^2\theta}{2}+C$.

Getting two different results for the same integral is something I can't grasp at all.

2. ## trigonometric identities

Originally Posted by Elvis
I'm trying to find the integral $\int\sec^2{\theta}\tan{\theta}\; {d\theta}$. Here is what I did:

Let $u = \tan{\theta}$. Then $\dfrac{du}{d\theta} = \sec^2\theta \Rightarrow {\color{blue}d{\theta} = \dfrac{du}{\sec^2\theta}}$ so that $\int\sec^2{\theta}\tan{\theta}\;{d\theta} = \int\dfrac{\sec^2{\theta}u}{sec^2\theta}\;{du} = \int{u}\;{du} = \frac{u^2}{2}+C$ $\Rightarrow \int\sec^2{\theta}\tan{\theta}\;{d\theta} = \frac{\tan^2\theta}{2}+C$, which is right.

My first question is, is there something wrong with the highlighted step? I have seen people raising questions about $\dfrac{dy}{dx}$ being treated as if they were numbers.

My second question is to do with when I tried to substitute $u$ for $\sec^2\theta$ and got different results than the above:

Let $u = \sec^2\theta$. Then $\dfrac{du}{d\theta} = 2\sec^2\theta\tan\theta \Rightarrow {\color{blue}d{\theta} = \dfrac{du}{2\sec^2\theta\tan\theta}}$ so that $\int\sec^2{\theta}\tan{\theta}\;{d\theta} = \int\dfrac{u\tan\theta}{2u\tan\theta}\;{du} = \frac{1}{2}\int\frac{u}{u}\;{du} = \frac{u}{2}+C$ $\; \Rightarrow \int\sec^2{\theta}\tan{\theta}\;{d\theta} = \frac{\sec^2\theta}{2}+C$.

Getting two different results for the same integral is something I can't grasp at all.
sec^2(theta)-tan^2(theta)=1
so when you substitute for any equation either of the values you will be left with a constant which goes with c.

3. Originally Posted by Pulock2009
sec^2(theta)-tan^2(theta)=1
so when you substitute for any equation either of the values you will be left with a constant which goes with c.
Could you please elaborate a bit further? I would lying if I say I understood what you have just said.

4. Originally Posted by Elvis
I'm trying to find the integral $\int\sec^2{\theta}\tan{\theta}\; {d\theta}$. Here is what I did:

Let $u = \tan{\theta}$. Then $\dfrac{du}{d\theta} = \sec^2\theta \Rightarrow {\color{blue}d{\theta} = \dfrac{du}{\sec^2\theta}}$ so that $\int\sec^2{\theta}\tan{\theta}\;{d\theta} = \int\dfrac{\sec^2{\theta}u}{sec^2\theta}\;{du} = \int{u}\;{du} = \frac{u^2}{2}+C$ $\Rightarrow \int\sec^2{\theta}\tan{\theta}\;{d\theta} = \frac{\tan^2\theta}{2}+C$, which is right.

My first question is, is there something wrong with the highlighted step? I have seen people raising questions about $\dfrac{dy}{dx}$ being treated as if they were numbers.

My second question is to do with when I tried to substitute $u$ for $\sec^2\theta$ and got different results than the above:

Let $u = \sec^2\theta$. Then $\dfrac{du}{d\theta} = 2\sec^2\theta\tan\theta \Rightarrow {\color{blue}d{\theta} = \dfrac{du}{2\sec^2\theta\tan\theta}}$ so that $\int\sec^2{\theta}\tan{\theta}\;{d\theta} = \int\dfrac{u\tan\theta}{2u\tan\theta}\;{du} = \frac{1}{2}\int\frac{u}{u}\;{du} = \frac{u}{2}+C$ $\; \Rightarrow \int\sec^2{\theta}\tan{\theta}\;{d\theta} = \frac{\sec^2\theta}{2}+C$.

Getting two different results for the same integral is something I can't grasp at all.
the two resulting antiderivatives differ only by a constant.

5. Originally Posted by Elvis
I'm trying to find the integral $\int\sec^2{\theta}\tan{\theta}\; {d\theta}$. Here is what I did:

Let $u = \tan{\theta}$. Then $\dfrac{du}{d\theta} = \sec^2\theta \Rightarrow {\color{blue}d{\theta} = \dfrac{du}{\sec^2\theta}}$ so that $\int\sec^2{\theta}\tan{\theta}\;{d\theta} = \int\dfrac{\sec^2{\theta}u}{sec^2\theta}\;{du} = \int{u}\;{du} = \frac{u^2}{2}+C$ $\Rightarrow \int\sec^2{\theta}\tan{\theta}\;{d\theta} = \frac{\tan^2\theta}{2}+C$, which is right.

My first question is, is there something wrong with the highlighted step? I have seen people raising questions about $\dfrac{dy}{dx}$ being treated as if they were numbers.

My second question is to do with when I tried to substitute $u$ for $\sec^2\theta$ and got different results than the above:

Let $u = \sec^2\theta$. Then $\dfrac{du}{d\theta} = 2\sec^2\theta\tan\theta \Rightarrow {\color{blue}d{\theta} = \dfrac{du}{2\sec^2\theta\tan\theta}}$ so that $\int\sec^2{\theta}\tan{\theta}\;{d\theta} = \int\dfrac{u\tan\theta}{2u\tan\theta}\;{du} = \frac{1}{2}\int\frac{u}{u}\;{du} = \frac{u}{2}+C$ $\; \Rightarrow \int\sec^2{\theta}\tan{\theta}\;{d\theta} = \frac{\sec^2\theta}{2}+C$.

Getting two different results for the same integral is something I can't grasp at all.
There is nothing wrong with the substitution you have made in the first operation.

I hope you have learned trigonometry and you know that

$sec^{2}x - tan^{2}x =1$

so,

your first answer is $\frac{\tan^2\theta}{2}+C$

the second answer you have is $\frac{\sec^2\theta}{2}+C$

= $\frac{(1+\tan^2\theta)}{2}+C$

= $\frac{tan^{2}\theta}{2}+\frac{1}{2}+C$

Note that $\frac{1}{2}+C= C'$ is also a constant. So it gives you $\frac{tan^{2}\theta}{2}+C'$