Hi all, just a quick question:(adsbygoogle = window.adsbygoogle || []).push({});

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 :)

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# B Integrating Problem

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