Solving Integral Equation: $\frac{\sin\left(x\right)}{\cos^{3}\left(x\right)}$

[Saracen Rue]

Hi all, just a quick question:

I'm trying to integrate this function in two different ways and I'm getting a different answer each way, can someone please quickly tell me where I'm going wrong? I've read through it for a couple hours and can't pick up the mistake.

$\int _{ }^{ }\frac{1}{2}\frac{\sin \left(x\right)}{\cos ^3\left(x\right)}dx=-\frac{1}{2}\int _{ }^{ }\frac{-\sin \left(x\right)}{\cos ^3\left(x\right)}dx=-\frac{1}{2}\int _{ }^{ }\left(\frac{1}{u^3}\right)\frac{du}{dx}\cdot dx=-\frac{1}{2}\int _{ }^{ }\left(\frac{1}{u^3}\right)du=-\frac{1}{2}\cdot \frac{-1}{2}\cdot \frac{1}{u^2}+c=\frac{1}{4\cos ^2\left(x\right)}+c$

$\int _{ }^{ }\frac{1}{2}\frac{\sin \left(x\right)}{\cos ^3\left(x\right)}dx=\frac{1}{2}\int _{ }^{ }\frac{\sin \left(x\right)}{\cos ^3\left(x\right)}dx=\frac{1}{2}\int _{ }^{ }\frac{\tan \left(x\right)}{\cos ^2\left(x\right)}dx=\frac{1}{2}\int _{ }^{ }u\cdot \frac{du}{dx}\cdot dx=\frac{1}{2}\int _{ }^{ }u\cdot du=\frac{1}{4}u^2+c=\frac{1}{4}\tan ^2\left(x\right)+c$

Thanks for your time :)

 

[Charles Link]

The only difference is the constants of integration will be different. Both are correct. $tan^2(\theta)+1 =sec^2(\theta)$.

 

[fresh_42]

Hi all, just a quick question:

I'm trying to integrate this function in two different ways and I'm getting a different answer each way, can someone please quickly tell me where I'm going wrong? I've read through it for a couple hours and can't pick up the mistake.

$\int _{ }^{ }\frac{1}{2}\frac{\sin \left(x\right)}{\cos ^3\left(x\right)}dx=-\frac{1}{2}\int _{ }^{ }\frac{-\sin \left(x\right)}{\cos ^3\left(x\right)}dx=-\frac{1}{2}\int _{ }^{ }\left(\frac{1}{u^3}\right)\frac{du}{dx}\cdot dx=-\frac{1}{2}\int _{ }^{ }\left(\frac{1}{u^3}\right)du=-\frac{1}{2}\cdot \frac{-1}{2}\cdot \frac{1}{u^2}+c=\frac{1}{4\cos ^2\left(x\right)}+c$

$\int _{ }^{ }\frac{1}{2}\frac{\sin \left(x\right)}{\cos ^3\left(x\right)}dx=\frac{1}{2}\int _{ }^{ }\frac{\sin \left(x\right)}{\cos ^3\left(x\right)}dx=\frac{1}{2}\int _{ }^{ }\frac{\tan \left(x\right)}{\cos ^2\left(x\right)}dx=\frac{1}{2}\int _{ }^{ }u\cdot \frac{du}{dx}\cdot dx=\frac{1}{2}\int _{ }^{ }u\cdot du=\frac{1}{4}u^2+c=\frac{1}{4}\tan ^2\left(x\right)+c$

Thanks for your time :)

That was a tough one. You've hidden the difference in the constant. The result is
$$
\frac{1}{4cos^2x}= \frac{sin^2 x + cos^2 x}{4cos^2x}= \frac{1}{4}(tan^2 x +1)
$$

Edit: @Charles Link beat me to it. I probably have searched too long where exactly the difference appears.

 

[FactChecker]

They only differ by a constant. In your first result, substitute the numerator 1 = sin²+cos².

Edit: And I am the slowest of all. :>)

 

[Saracen Rue]

The only difference is the constants of integration will be different. Both are correct. $tan^2(\theta)+1 =sec^2(\theta)$.

That was a tough one. You've hidden the difference in the constant. The result is
$$
\frac{1}{4cos^2x}= \frac{sin^2 x + cos^2 x}{4cos^2x}= \frac{1}{4}(tan^2 x +1)
$$

Edit: @Charles Link beat me to it. I probably have searched too long where exactly the difference appears.

They only differ by a constant. In your first result, substitute the numerator 1 = sin²+cos².

Edit: And I am the slowest of all. :>)

Thanks everyone :) I knew it would be something little like that. I'm always fine with applying the deriving/integration rules for particular functions and whatnot, but I typically end up messing up basic algebra haha. Anyway, thank you all again :)