Solve the intergral ?

Question

(i) Use trigonometric substitution to evaluate the intergral:

#int_0^sqrt(2)##(dx)/((x^2+2)^2#

(ii) Solve the integral equation

#int_1^x# #f(t) dt = f(x) - 5#
where #x>=##1# and express your answer as an exponential.

4801 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-04-11T20:58:33

First problem:
Guessing that we should try this transform: #x = sqrt 2tan psi# and #tan psi = x/sqrt 2#

This leaves:

#int_0^(sqrt 2) \ (dx)/((x^2+2)^2#

#= int_0^(pi/4) \ (d(sqrt 2 tan psi))/((2 tan^2 psi+2)^2#

#= sqrt 2/(4) int_0^(pi/4) \ d psi sec^2 psi 1/((sec^2 psi)^2#

#= sqrt 2/(4) int_0^(pi/4) \ d psi cos^2 psi #

#= sqrt 2/(8) int_0^(pi/4) \ \ d psi cos2 psi + 1 #

#= sqrt 2/(8) [ (sin2 psi)/2 + psi]_0^(pi/4) #

#= sqrt 2/(8) [ {1/2 + pi/4} - {0} ] #

#= sqrt 2/(16) [ 1 + pi/2] #

The second problem:

#int_1^x f(t) dt = f(x) - 5#

Using Liebnitz's Rule:

#d/dx( int_1^x f(t) dt = f(x) - 5) implies f(x) = f'(x)#

Which solves as:

#f(x) = alpha e^x#

With this theme:

#int_1^x alpha e^t dt = alpha(e^x - e) #

#implies alpha e^x - alpha e = alpha e^x - 5# or #alpha= 5/e#

#implies f(x) = 5 e^(x-1)#