Mathematics Test - 6

Hence equation of tangent to the given curve at (0 , 1) is :

(y −1) = 2(x −0),i.e..y −1 = 2x  

Begin by rewriting  ∫tan⁴ xdx  as  ∫tan² xtan² xdx.    

Now we can apply the Pythagorean Identity, tan² x+1=sec² x,                     or  tan² x=sec² x −1

∫tan² x tan² x dx = ∫(sec² x −1)tan² xdx

Distributing the  tan² x:
       = ∫sec² xtan² x  −tan² xdx
Applying  the  sum rule:
        = ∫sec² xtan² xdx −∫tan² xdx

We 'll evaluate these integrals one by one.

First Integral  
This one is solved using a  
Let  u = tanx


Applying the substitution,
Because  u = tanx,

Second Integral
Since we don 't really know what  ∫tan² xdx  is by just looking at it, try applying the  tan² x  = sec² x −1  

identity again:

∫tan² xdx = ∫(sec² x −1)dx

Using the sum rule, the integral boils down to:
∫sec² xdx −∫1dx

The first of these, ∫sec² xdx, is just  tanx + C.

The second one, the so-called "perfect integral ", is simply  x+C.

Putting it all together, we can say:
∫tan² xdx = tanx + C −x + C

And because  C+C  is just another arbitrary constant, we can combine it into a general constant  C:
∫tan² xdx = tanx −x + C

Combining the two results, we have:
∫tan⁴ xdx=∫sec² xtan² xdx −∫tan² xdx

=(tan³ x/3 + C) −(tanx −x + C)

=tan³ x/3 −tanx + x + C

Again, because  C+C  is a constant, we can join them into one  C.

∫(2+tan x)² dx
= ∫(4 + tan² x + 4tan x)dx
= ∫4 dx + ∫tan² x dx + 4 ∫tan x dx
= 4x + ∫(sec² x - 1)dx + 4(log|sec x|)
= 4x + tanx - x + 4(log|sec x|)
3x + tanx + 4(log|sec x|) + c

Correct Answer : d

Explanation : ∫(-1 to 1) e^|x| dx

∫(-1 to 0) e^|x| dx + ∫(0 to 1) e^|x| dx

 e¹ -1 + e¹ - 1

=>2(e - 1)

 ∫(-a to a)f(x)dx
= ∫(0 to a) [f(x) + f(-x)] if f(x) is an odd function
⇒f(-x) = -f(x)