# Evaluate the integral using the Fundamental Theorem of Calculus.

$$\int_{\frac{-\pi}{4}}^{0} (\sec x \tan x) dx$$

I need to evaluate the integral using the Fundamental Theorem of Calculus.

-
$\sec x\tan x$ is the derivative of $\sec x$. –  André Nicolas Feb 2 '13 at 6:55

Let $u=\sec(x)$. Then $du = \sec(x) \tan(x) dx$. We see this is just a perfect differential. So, by the fundamental theorem, the integral is $\sec(0) - \sec(\frac{-\pi}{4}) = 1 - \sqrt{2}$

-
I did it this way. Thank you for the help. I didnt even think to use U-substitution. –  Ak47 Feb 2 '13 at 7:08

Hint: $\sec x \tan x$ is the derivative of $\sec x$.

One way to see this is to rewrite $\sec x$ and $\tan x$ in terms of sine and cosine. We get $\dfrac{\sin x}{\cos^2 x}$. Make the subsitution $u=\cos x$.

Remark: One version of the Fundamental Theorem of Calculus says that if $f(x)$ is a well-behaved function on the interval $[a,b]$, then $$\int_a^b f(x)\,dx=F(b)-F(a),$$ where $F(x)$ is any function whose derivative is $f(x)$,

-

$\int_{\frac{-\pi}{4}}^0{\sec x\tan xdx}=\sec x|_{\frac{-\pi}{4}}^{0}= \sec(0)-\sec(\frac{-\pi}{4})=1-\sqrt{2}$

-
Note on my edit: \sec and \tan are predefined operators in Latex. –  Michael Albanese Feb 2 '13 at 7:06
ohh okey thank you :) i don't know about that :p –  A Ricko Maulidar Feb 2 '13 at 7:07