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I am trying to find the integral of $\int \tan x \sec^3 x dx$

$\int \tan x(1+\tan^2 x)\sec x\, dx$

This gets me nowhere since I get a $\sec^2 x$ derivative with tan substitution so I try something else.

$\int \frac {\sin x}{\cos x} \frac{1}{\cos^3x} dx$

$\int \frac {\sin x}{\cos^4 x} dx$

$u = \cos x$ $du = -\sin x$

$\int \frac {-1}{u^4} du$

$-1\int {u^{-4}} du$

$-1 \frac{u^{-3}}{3}$ $\frac{-1}{3\cos^3 x}$

This for some reason is wrong.

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    Another method would be to note that $\tan x \, \sec^3 x = (\tan x \, \sec x)(\sec^2 x)$ and that the differential of $\sec x$ is $\tan x \, \sec x$. A method called spotting the presence of the derivative.2012-06-03

2 Answers 2

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You missed a $-$ sign in the denominator. $$\int x^{n} \ \text dx = \frac{x^{n+1}}{n+1} + C$$ when $n \neq -1$.

So when $n=-4$ you get $$\int x^{-4} \ \text dx = -\frac{1}{3} \cdot x^{-3} +C$$

There is still an easier method of doing this: $$\int \tan{x}\cdot \sec^{3}(x) \ \text dx = \int \tan{x} \cdot \sec{x} \cdot \sec^{2}(x) \ \text dx = \int t^{2} \ \text dt$$ by putting $t=\sec{x}$.

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    @jordan: Great that you were able to come up with a solution on your own. Here is an exercise : $\int\frac{1}{\sqrt{\sin{x} \cdot \cos{x}}} \ dx$2012-06-03