Reconhecendo integrações de funções trigonometricas

Before we start, I assume you're familiar with the derivatives of the six trigonometric functions: $$\dfrac{d}{dx} \sin{x} = \cos{x} \\ \dfrac{d}{dx} \cos{x} = -\sin{x} \\ \dfrac{d}{dx} \tan{x} = \sec^2{x} \\ \dfrac{d}{dx} \csc{x} = -\csc{x}\cot{x} \\ \dfrac{d}{dx} \sec{x} = \tan(x)\sec{x} \\ \dfrac{d}{dx} \cot{x} = -\csc^2{x}$$

Via integrating these formulas, we can immediately deduce that $$\int{\sin{x}} \ dx= -\cos{x} \\ \int{\cos{x}} \ dx= \sin{x} \\ \int{\sec^2{x}} \ dx= \tan{x} \\ \int{\csc{x}\cot{x}} \ dx= -\csc{x} \\ \int{\tan{x}\sec{x}} \ dx= \sec{x} \\ \int{\csc^2{x}} \ dx= -\cot{x}$$

Now let's try and find a formula for $\int{\tan{x}} \ dx$. This will require knowledge of U-substitution.

What is $\tan(x)$, really? It's just a ratio of sine to cosine. So we can rewrite our integral as $$\int{\tan(x)} \ dx = \int{\dfrac{\sin(x)}{\cos(x)}} \ dx$$

Sines and cosines in a fraction... this sounds like a good substitution! If we let $u = \cos{x}$, we'll have $du = -\sin{x} \ dx$. Substituting this in: $$\begin{align*} \int{\dfrac{\sin(x)}{\cos(x)}} \ dx &= -\int{\dfrac{du}{u}} \\ &= -\ln{|u|} + C \\ &= \ln{(|\cos{x}|^{-1})} + C \\ \int{\tan(x)} \ dx &= \ln{|\sec{x}|} + C \end{align*}$$

Now we have an expression for $\int{\tan{x}} \ dx$.

Using the same principle, try to see if you can find $\int{\cot{x}} \ dx$.

If you did everything correctly, you should have gotten that $\int{\cot{x}} \ dx = \ln{\sin{x}} + C$. Well done!

Now that you have learned how to do the integrals of all 6 trigonometric functions, you can now do ALL the problems on here, congratulations! Best of luck on your new adventure!