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Integrate the function.

$$ \int \sec^2 x \tan x dx $$

I'm trying to get a proper substitution, but I couldn't get anything proper.

hardmath
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3 Answers3

2

We know that, $$ d(\tan(x)) = \sec^2(x).dx$$ Plugging the LHS of the above equation in place of RHS into the main expression gives: $$ \int \tan x. d( \tan x )$$ $$ \frac{\tan^2(x)}{2} + Constant$$

lsp
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2

This is a good one. Hope you are familiar with integration by substitution.

now case 1:

put $\\ \tan x=t$ so, $ dt=\sec^2xdx $ Substituting, we get

$I= \int tdt\Rightarrow \frac {t^2}{2}+c_1 $

that is, I= $ \frac{\tan^2x}{2} + c_1 $

Case 2: put $ \sec x=u $

$du=\sec x\tan x \ dx$

so, $I= \int udu => I=\frac {u^2}{2}+c_2$

since both the integrals are equal,

$\frac{\tan^2x}{2}+c_1=\frac {\sec^2x}{2}+c_2 $

which implies $c_1 -c_2 =\frac {1}{2} $.

So, we understand that, if a function has more than one integral, the integrals differ by a constant.

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dajoker
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  • A function can have more than one integral? I didnt know that.. –  Jun 07 '13 at 09:09
  • This has "more than one integral" only because of (in this case) the trigonometric identities that link $\tan$ and $\sec$. The two are identical for all values of $x$. Another example would be something that resulted in $x^2-a^2$, but this would also be the same as $(x-a)(x+a)$, by virtue of an algebraic (rather than trigonometric) identity. – danimal Mar 16 '15 at 12:28
1

Substitute $u=\tan x$, then you'll get $du=\sec^2(x) dx$

$$\int \sec^2(x) \cdot \tan x dx = \int u \; du = \frac{u^2}{2}+c \Rightarrow \frac{\tan^2(x)}{2}+c$$

Now we use $\sec^2(x)=1+\tan^2(x)$ and finally

$$\int \sec^2x \cdot \tan x dx = \frac{\sec^2(x)}{2}+c$$

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