Ejercicios de cálculo de primitivas de funciones trigonométricas

ENUNCIADO. Calcúlense las siguientes integrales indefinidas:
a) $\displaystyle \int\,\dfrac{1}{\cos\,x}\,dx$
b) $\displaystyle \int\, \dfrac{1}{\cos^3\,x}\,dx$

SOLUCIÓN.
a)
$\displaystyle \int\,\dfrac{1}{\cos\,x}\,dx=$
  $\displaystyle =\int\,\sec\,x\,dx$
    $\displaystyle =\int\,\sec\,x \cdot \dfrac{\sec\,x+\tan\,x}{\sec\,x+\tan\,x}\,dx$
      $\displaystyle =\int\,\dfrac{\sec^2\,x + \sec\,x\cdot \tan\,x}{\sec\,x+\tan\,x}\,dx$
        $\displaystyle \overset{(1)}{=}\int\,\dfrac{\sec^2\,x + \sec\,x\cdot \tan\,x}{\sec\,x+\tan\,x}\,dx$
          $\displaystyle = \int\,\dfrac{du}{u}$
            $\displaystyle = \ln\,|u|+C$
              $\displaystyle = \ln\,|\sec\,x+\tan\,x|+C$
                $\displaystyle = \ln\,\left|\dfrac{1}{\cos\,x}+\tan\,x\right|+C \quad \quad (I)$

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(1) Cambio de variable: $u=\sec\,x+\tan\,x \Rightarrow du = \dfrac{1}{\cos^2\,x}\, ( 1+\sin\,x)\,dx = (\sec^2\,x + \tan\,x\cdot \sec\,x)\,dx$
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-oOo-
b)
$\displaystyle \int\, \dfrac{1}{\cos^3\,x}\,dx=$
  $=\displaystyle \int\, \dfrac{1}{\cos^2\,x}\cdot \dfrac{1}{\cos\,x}\,dx$
    $\overset{(2)}{=}\displaystyle \dfrac{\tan\,x}{\cos\,x}-\int\, \dfrac{\tan\,x \cdot \sin\,x}{\cos^2\,x}\,dx$
      $=\displaystyle \dfrac{\tan\,x}{\cos\,x}-\int\, \dfrac{\sin^2\,x}{\cos^3\,x}\,dx$
        $=\displaystyle \dfrac{\tan\,x}{\cos\,x}-\int\, \dfrac{1-\cos^2\,x}{\cos^3\,x}\,dx$
          $=\displaystyle \dfrac{\tan\,x}{\cos\,x}-\int\, \dfrac{1}{\cos^3\,x}\,dx+\int\, \dfrac{\cos^2\,x}{\cos^3\,x}\,dx$
            $=\displaystyle \dfrac{\tan\,x}{\cos\,x}-\int\, \dfrac{1}{\cos^3\,x}\,dx+\int\, \dfrac{1}{\cos\,x}\,dx$
Con lo cual
$$2\,\displaystyle \int\, \dfrac{1}{\cos^3\,x}\,dx=\displaystyle \dfrac{\tan\,x}{\cos\,x}+\int\, \dfrac{1}{\cos\,x}\,dx$$
luego, teniendo en cuenta (I) el resultado del apartado anterior
$$2\,\displaystyle \int\, \dfrac{1}{\cos^3\,x}\,dx=\displaystyle \dfrac{\tan\,x}{\cos\,x}+\ln\,\left|\dfrac{1}{\cos\,x}+\tan\,x\right|$$ de donde se deduce que
$$\displaystyle \int\, \dfrac{1}{\cos^3\,x}\,dx=\dfrac{1}{2}\left( \dfrac{\tan\,x}{\cos\,x}+\ln\,\left|\dfrac{1}{\cos\,x}+\tan\,x\right| \right)+C$$

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(2) Por partes:
$w=\dfrac{1}{\cos\,x} \Rightarrow dw = -\dfrac{1}{\cos^2\,x}\cdot ( -\sin\,x)\,dx = \dfrac{\sin\,x}{\cos^2\,x}\,dx$
$dt=\dfrac{1}{\cos^2\,x}\,dx \Rightarrow t=\tan\,x+k$
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$\square$