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Evaluate $\displaystyle \int \tan^2x\sec^2x\,dx$

I tried several methods:

  • First method was I changed $\tan^2x = \sec^2x-1$, and then substitute $\sec x$ to $t$, but it doesn't work.
  • Second method was to use substitute $\tan^2x = v$, $\sec x = u$. And, it does not work as well.

Is there any better way to solve this problem?

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2 Answers 2

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Substituting $x = \tan^{-1}t$ and using $\sec^2(x)= \tan^2(x)+1$ we get $$\int (1+t^2)\cdot t^2\cdot \frac{1}{1+t^2}\, dt = \int t^2dt = \frac{t^3}{3}+C.$$ Since $t=\tan(x)$, we recover $$\int \sec^2(x) \cdot \tan^2(x)\, dx = \frac{\tan^3(x)}{3}+C.$$

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$$I=\int\sec x\tan^2xdx\\=\int\sec x(\sec^2x-1)dx\\=\int\underbrace{\sec x}_{\mathtt u}.\underbrace{\sec^2x}_{\mathtt v}dx-\int\sec xdx\\=\left(\sec x\int\sec^2 xdx-\int (\sec x\tan x)\tan xdx\right)-\int\sec x$$ $$I=\sec x \tan x-\ln|\sec x+\tan x|-I+C$$ $$I=\frac12\left(\sec x\tan x-\ln\left|\sec x+\tan x\right|\right)+C$$

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  • $\begingroup$ Thanks for taking care of detail :) I appreciate your efforts! I understood what you wrote. Thank you again :D $\endgroup$
    – Jason
    Sep 19, 2014 at 9:24

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