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# Properties of Definite Integrals - III

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#### Properties of Definite Integrals - III - Lesson Summary

Property 6:

âˆ«02a f(x) dx = âˆ«0a f(x) dx + âˆ«0a f(2a - x) dx

Proof:

âˆ«02a f(x) dx

By a property on definite integrals,

if a < c < b , then âˆ«ab f(x) dx = âˆ«ac f(x) dx + âˆ«cb f(x) dx

âˆ«02a f(x) dx = âˆ«0a f(x) dx + âˆ«a2a f(x) dx -----(1)

Let's evaluate âˆ«02a f(x) dx

Let t = 2a - x â‡’ x = 2a - t

dx = 0 - dt â‡’ dx = -dt

When x = a, t = 2a - a = a

When x = 2a, t = 2a - 2a = 0

âˆ«02a f(x) dx = âˆ«a0 f(2a - t) x (-dt)

= - âˆ«a0 f(2a - t) dt

By a property of definite integrals,

âˆ«ab f(x) dx = - âˆ«ba f(x) dx

= âˆ«0a f(2a - t) dt

By a property of definite integrals,

âˆ«ab f(x) dx = âˆ«ab f(t) dt

=  âˆ«0a f(2a - x) dx

âˆ´ âˆ«02a f(x) dx = âˆ«0a f(2a - x) dx ----(2)

On substituting equation 2, in equation 1, we get

âˆ«02a f(x) dx = âˆ«0a f(x) dx + âˆ«0a f(2a - x) dx

Example:

Evaluate âˆ«0Ï€  (x sin x)/(1 + cos2x) dx.

Let I = âˆ«0Ï€  (x sin x)/(1 + cos2x) dx

= âˆ«02Ï€/2  (x sin x)/(1 + cos2x) dx

âˆ«02a f(x) dx = âˆ«0a f(x) dx + âˆ«0a f(2a - x) dx

= âˆ«02×Ï€/2  (x sin x)/(1 + cos2x) dx

= = âˆ«0Ï€/2  (x sin x)/(1 + cos2x) dx +  âˆ«0Ï€/2  ((Ï€- x) sin (Ï€-x))/(1 + cos2(Ï€-x)) dx

= âˆ«0Ï€/2  (x sin x)/(1 + cos2x) dx +  âˆ«0Ï€/2  ((Ï€- x) sin x)/(1 + (- cos x )2) dx

= âˆ«0Ï€/2  (x sin x)/(1 + cos2x) dx +  âˆ«0Ï€/2  (Ï€ sin x - x sin x)/(1 + cos2 x) dx

= âˆ«0Ï€/2  (x sin x + Ï€ sin x - x sin x)/(1 + cos2x) dx

= âˆ«0Ï€/2  (Ï€ sin x)/(1 + cos2x) dx

= Ï€ âˆ«0Ï€/2  sin x/(1 + cos2x) dx -----(1)

Let cos x = t

â‡’ - sin x dx = dt

â‡’ sin x dx = - dt

When x = Ï€/2 , t = cos Ï€/2 = 0

When x = 0 , t = cos 0 = 1

= Ï€ âˆ«10 -1/(1+t2) dt

= -Ï€ âˆ«10 dt/(1+t2)

âˆ«ab f(x) dx = - âˆ«ba f(x) dx

-Ï€ âˆ«10 dt/(1+t2) = = Ï€ âˆ«01 dt/(1+t2)

âˆ« 1/(1+t2) dt = tan-1t + C

Ï€ âˆ« dt/(1+t2) = Ï€ [tan-1t]01

= Ï€ [tan-11 - tan-10]

= Ï€(Ï€/4 - 0)

= Ï€2/4

Property 7:

âˆ« 0 2a f(x) = 2 âˆ« 0 a f(x)dx, if f(2a -x) = f(x) 0, if f(2a - x) = -f(x)

Proof:

âˆ«02a f(x) dx = âˆ«0a f(x) dx + âˆ«0a f(2a - x) dx----(1)

If f(2a - x) = f(x),   then from (1) , we have

âˆ«02a f(x) dx = âˆ«0a f(x) dx + âˆ«0a f(x) dx

= 2 . âˆ«0a f(x) dx

If f(2a - x) = - f(x), then from (1), we have

âˆ«02a f(x) dx = âˆ«0a f(x) dx - âˆ«0a f(x) dx

= 0

âˆ« 0 2a f(x) = 2 âˆ« 0 a f(x)dx, if f(2a -x) = f(x) 0, if f(2a - x) = -f(x)

Example:

Evaluate âˆ«0Ï€ tan x/(sec x + cos x) dx

Let f(x) = tan x/(sec x + cos x)  and 2a = Ï€

Let's find out f(2a - x) = f(Ï€-x) = tan (Ï€-x)/(sec (Ï€-x) + cos (Ï€-x))

= - tanx/(-secx-cosx)

= tanx/(secx + cosx)

= f(x)

By a property on definite integrals,

âˆ«02a f(x) dx = 2. âˆ«0a f(x) dx if f(2a - x) = f(x)

âˆ´ âˆ«0Ï€ tanx/(secx + cosx) dx = 2 âˆ«0Ï€/2 tanx/(secx + cosx) dx

= 2 âˆ«0Ï€/2 (sin x/cox x)/(1/cos x + cosx) dx

= 2 âˆ«0Ï€/2 sin x/(1 + cos2x) dx

Let cos x = t

On differentiating both sides, we get

- sin x dx = dt

sin x dx = -dt

When x = 0 , t = cos 0 = 1

When x = Ï€/2; t = cos Ï€/2 = 0

Therefore, 2 âˆ«0Ï€/2 sin x/(1 + cos2x) dx = 2.âˆ«10 -dt/(1+t2)

âˆ«ab f(x) dx = - âˆ«ba f(x) dx

2 .âˆ«10 -dt/(1+t2) = 2 .âˆ«01 1/(1+t2) dt

= 2 [tan-1 t]01

= 2[tan-1(1) - tan-1(0)]

= 2[Ï€/4 - 0]

= Ï€/2

âˆ«0Ï€ tanx/(sec x + cos x) dx = = Ï€/2

Property 8:

i. âˆ«-aa f(x) dx = 2 âˆ«0a f(x) dx,

if f is an even function , i.e, f(-x) = f(x)

ii. âˆ«-aa f(x) dx = 0

if f is an odd function , i.e, f(-x) = - f(x)

Proof:

i. LHS = âˆ«-aa f(x) dx

If a < c < b, then

âˆ«ab f(x) dx = âˆ«ac f(x) dx + âˆ«cb f(x) dx

We have - a < 0 < a and

âˆ«-aa f(x) dx = âˆ«-a0 f(x) dx + âˆ«0a f(x) dx  ------(1)

âˆ«-a0 f(x) dx

Let x = -t

dx = -dt

When x = -a, then -t = a â‡’ t = a

When x = 0 , then -t = 0 â‡’ t = 0

âˆ«-a0 f(x) dx = âˆ«-a0 f(-t) (-dt)

= - âˆ«a0 f(-t) dt

= âˆ«0a f(-t) dt      ['.' âˆ«ab f(x) dx = - âˆ«ba f(x) dx]

âˆ«-a0 f(x) dx = âˆ«0a f(-t) dt ----(2)

However, by a property of definite integrals,

âˆ«ab f(x) dx = âˆ«ab f(t) dt

âˆ«-a0 f(x) dx = âˆ«0a f(-x) dx -----(3)

âˆ«-aa f(x) dx = âˆ«0a f(-x) dx + âˆ«0a f(x) dx -----(4)

Case - 1:If the function is an even function, i.e.

f(-x) = f(x):

âˆ«-aa f(x) dx = âˆ«0a f(-x) dx + âˆ«0a f(x) dx

â‡’ âˆ«-aa f(x) dx = âˆ«0a f(x) dx + âˆ«0a f(x) dx

â‡’ âˆ«-aa f(x) dx = 2 âˆ«0a f(x) dx

Case - 2: If the function is an odd function, i.e.

f(-x) = - f(x):

â‡’ âˆ«-aa f(x) dx = âˆ«0a f(-x) dx + âˆ«0a f(x) dx

â‡’ âˆ«-aa f(x) dx = âˆ«0a -f(x) dx + âˆ«0a f(x) dx

â‡’ âˆ«-aa f(x) dx = - âˆ«0a f(x) dx + âˆ«0a f(x) dx

â‡’ âˆ«-aa f(x) dx = 0

i. âˆ«-aa f(x) dx = 2. âˆ«0a f(x) dx,

if f is an even function , i.e f(-x) = f(x)

ii. âˆ«-aa f(x) dx =  0

if f is an odd function , i.e f(-x) = -f(x)

Example:

Evaluate âˆ«log½log2 sin ((ex-1)/(ex+1))dx

âˆ«log½log2 sin ((ex-1)/(ex+1))dx = âˆ«-log2log2 sin ((ex-1)/(ex+1))dx

Let f(x) = sin ((ex-1)/(ex+1))

Now, f(-x) = sin ((e-x-1)/(e-x+1))

= sin ((e1/x-1)/(e1/x+1))

= sin ((1 - ex)/(1+ex))

= sin (-(ex-1)/(1+ex))

= - sin ((ex-1)/(ex+1)) ['.' sin(-x) = - sin x]

= - f(x)

â‡’ f(-x) = - f(x)

âˆ´ f(x) is an odd function

âˆ«-aa f(x) dx =  0

if f is an odd function , i.e, f(-x) = - f(x)

âˆ«log½log2 sin ((ex-1)/(ex+1))dx = 0