## Integration Worksheet5 solution2

In this page integration worksheet5 solution2 we are going to see solution of some practice question from the worksheet of integration.

Question 4

Integrate the following with respect to x,x/√(x² + 3)

Solution:

We are going to solve this problem by using substitution method. For that let us consider "t" as "(x² + 3)"

t = (x² + 3)

differentiate with respect to x

dt = (2 x + 0) dx

dt = 2 x dx

x dx = dt/2

= ∫ x/√(x² + 3) dx

= ∫ (dt/2)√t

= (1/2) ∫√t dt

= (1/2) ∫t^(1/2) dt

= (1/2) (t^[(1/2) + 1]/[(1/2) + 1]) + C

= (1/2) (t^(3/2)/(3/2) + C

= (1/2) (2/3) t^(3/2) + C

= (1/3) (x² + 3)^(3/2) + C

Question 5

Integrate the following with respect to x,(2 x + 3) √(x² + 3 x - 5)

Solution:

We are going to solve this problem by using substitution method. For that let us consider "t" as "(x² + 3 x - 5)"

t = x² + 3 x - 5

differentiate with respect to x

dt = (2 x + 3(1) - 0) dx

dt = (2 x + 3) dx

= ∫(2 x + 3) √(x² + 3 x - 5) dx

= ∫ (dt/√t)

= ∫t^(-1/2) dt

= t^[(-1/2) + 1]/[(-1/2) + 1] + C

= t^(1/2)/(1/2) + C

= 2 √(x² + 3 x - 5) + C

Question 6

Integrate the following with respect to x, tan x

Solution:

= ∫ tan x dx

= ∫ (sin x/cos x) dx

now we are going to consider t as "cos x"

t = cos x

differentiating with respect to x,we get

dt = -sin x dx

sin x dx = - dt

= ∫(- dt/t)

= - ∫(1/t) dt

= - log t + C

= - log (cos x) + C

Question 7

Integrate the following with respect to x, sec x

Solution:

= ∫ sec x dx

now we are going to multiply the numerator and denominator by (secx+tanx)

= ∫ [sec x (sec x + tan x)/(sec x + tan x)] dx

= ∫ [(sec ² x + sec x tan x)/(sec x + tan x)] dx

now we are going to consider the denominator as "t"

t = sec x + tan x

dt = sec x tan x + sec ² x

= ∫ dt/t

= ∫ (1/t) dt

= log t + C

= log (sec x + tan x) + C

integration worksheet5 solution2 integration worksheet5 solution2