Test: Indefinite Integration-Integration of General Functions(21 Sep) - JEE MCQ

= tanx + cotx + C

Begin by rewriting  ∫tan⁴xdx as ∫tan²xtan²xdx.      

Now we can apply the Pythagorean Identity,  tan²x+1=sec²x,                                         or tan²x=sec²x−1

∫tan²x tan²x dx = ∫(sec²x−1)tan²xdx

Distributing the tan²x:
             = ∫sec²xtan²x − tan²xdx
Applying the sum rule:
                = ∫sec²xtan²xdx − ∫tan²xdx

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  ∫tan²xdx is by just looking at it, try applying the tan²x = sec²x−1 

identity again:

∫tan²xdx = ∫(sec²x−1)dx

Using the sum rule, the integral boils down to:
∫sec²xdx − ∫1dx

The first of these,  ∫sec²xdx, is just tanx + C.

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

Putting it all together, we can say:
∫tan²xdx = tanx + C − x + C

And because C+C is just another arbitrary constant, we can combine it into a general constant C:
∫tan²xdx = tanx − x + C

Combining the two results, we have:
∫tan⁴xdx=∫sec²xtan²xdx−∫tan²xdx

=(tan³x/3 + C) − (tanx − x + C)

=tan³x/3 − tanx + x + C

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

∫(2+tan x)²dx
= ∫(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

 ∫sin²(2x+1) dx
Put t = 2x+1
dt = 2dx
dx= dt/2
= 1/2∫sin² t dt
=1/2∫(1-cos²t)/2 dt
= 1/4∫(1-cos²t) dt
= ¼[(t - (sin²t)/2]dt
= t/4 - sin2t/8 + c
= (2x+1)/4 - ⅛(sin(4x+2)) + c
= x/2 - 1/8sin(4x+2) + ¼ + c
As ¼ is also a constant, so eq is = x/2 - 1/8sin(4x+2) + c

 Let cos⁻¹(sinx)=θ
⇒ sinx=cosθ
⇒ sinx=sin(π/2−θ)
⇒ x = π/2−θ
⇒ θ = π/2−x
∴ ∫cos⁻¹(sinx)dx=∫(π/2−x)dx
= ∫π/2dx−∫xdx
= πx/2 - x²/2 + c, where b is a constant of integration.

 I = ∫cos2x/(sinx+cosx)²dx
⇒I = ∫cos²x−sin²x(sinx+cosx)²dx
⇒I = ∫[(cosx+sinx)(cosx−sinx)]/(sinx+cosx)²dx
⇒I = ∫(cosx−sinx)/(sinx+cosx)dx
Let sinx+cosx = t 
(cosx−sinx)dx = dt
Then, I = ∫dt/t
I = log|t|+c
I = log|sinx + cosx| + c