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Question

Evaluate the given integral.
$$\int { \sec ^{ 2 }{ \left( 7-4x \right)  }  } dx\quad $$


Solution

Let $$t=7-4x\Rightarrow\,dt=-4\,dx$$

$$\Rightarrow\,\dfrac{-dt}{4}=dx$$

$$I=\displaystyle\int{{\sec}^{2}{\left(7-4x\right)}dx}$$

$$=\dfrac{-1}{4}\displaystyle\int{{\sec}^{2}{t}dt}$$

Let $$u=\tan{t}\Rightarrow\,dt={\sec}^{2}{dt}$$

$$=\dfrac{-1}{4}\displaystyle\int{du}$$

$$=\dfrac{-1}{4}u+c$$       ..........where $$c$$ is the constant of integration.

$$=\dfrac{-1}{4}\tan{t}+c$$            ............where $$u=\tan{t}$$

$$=\dfrac{-1}{4}\tan{\left(7-4x\right)}+c$$          ............where $$t=7-4x$$

Mathematics

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