MHB 1.1.4 AP Calculus Exam Problem int sec x tan x dx

[karush]

$\tiny{213(DOY)}$

$\displaystyle\int \sec x \tan x \: dx =$

(A) $\sec x + C$

(B) $\tan x + C$

(C) $\dfrac{\sec^2 x}{2}+ C$

(D) $\dfrac {tan^2 x}{2}+C $

(E) $\dfrac{\sec^2 x \tan^2 x }{2}+ C$

Spoiler

$(\sec x)'=\sec x \tan x$ so the answer is $\sec x +C$ (A)

 

[topsquark]

213(DOY)
$\displaystyle\int \sec x \tan x \: dx =$

(A) $\sec x + C$

(B) $\tan x + C$

(C) $\dfrac{\sec^2 x}{2}+ C$

(D) $\dfrac {tan^2 x}{2}+C $

(E) ${\sec^2 x \tan^2 x }{2}+ C$

Spoiler

$(\sec x)'=\sec x \tan x$ so the answer is $\sec x +C$ (A)

Yup! (Muscle)

-Dan

 

[Olinguito]

A neat trick to do is to differentiate each of the multiple-choice answers in turn until you get the expression to be integrated.

 

[HallsofIvy]

Instead of memorizing a "secant-tangent" integral rule, I would use the fact that $tan(x)= \frac{sin(x)}{cos(x)}$ and $sec(x)= \frac{1}{cos(x)}$ so that $\int sec(x)tan(x)dx= \int \frac{sin(x)}{cos^2(x)}dx$. Let u= cos(x) so that du= -sin(x)dx and the integral becomes $-\int\frac{du}{u^2}= -\int u^{-2}du= -(-u^{-1})+ C= \frac{1}{cos(x)}+ C= sec(x)+ C$.

 

[karush]

Instead of memorizing a "secant-tangent" integral rule, I would use the fact that $tan(x)= \frac{sin(x)}{cos(x)}$ and $sec(x)= \frac{1}{cos(x)}$ so that $\int sec(x)tan(x)dx= \int \frac{sin(x)}{cos^2(x)}dx$. Let u= cos(x) so that du= -sin(x)dx and the integral becomes $-\int\frac{du}{u^2}= -\int u^{-2}du= -(-u^{-1})+ C= \frac{1}{cos(x)}+ C= sec(x)+ C$.

mahalo

btw how would you rate this problem :
easy, medium, hard

 

[HOI]

I would consider it about "medium-easy".

 

[karush]

how about 'medsy'