Mathematics

Integrate $$\int x{\sec ^2}x dx$$


SOLUTION

Consider given the given intigration,

Let,

  $$ I=\int{x.{{\sec }^{2}}xdx} $$

 $$ \because \int{u.vdx=u\int{vdx-\int{\left( \dfrac{du}{dx}\int{vdx} \right)dx}}} $$

 $$ \therefore \int{x.{{\sec }^{2}}xdx}=x\int{{{\sec }^{2}}xdx}-\int{\left( \dfrac{dx}{dx}\int{{{\sec }^{2}}x} \right)}dx $$

 $$ =x\tan x-\int{1.\tan x}dx $$

 $$ =x\tan x-\log \sec x+C $$


Hence, this is the answer.

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Subjective Medium Published on 17th 09, 2020
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