Integration Worksheet6 solution8

In this page integration worksheet6 solution8 we are going to see solution of some practice question from the worksheet of integration.

Question 18

Integrate the following with respect to x,  e^2x sin 3x

Solution:

∫ e^2x sin 3x dx

u = sin 3x              dv = e^2x

du = 3 cos 3 x        v = e^2x/2

∫ u dv = u v - ∫ v du

= (sin 3x)(e^2x/2) - ∫(e^2x/2) (3 cos 3 x) dx

= (sin 3 x)(e^2x/2) - (3/2) ∫ e^2x (cos 3x) dx--------(1)

now we are going to find the integration value of ∫ e^2x (cos 3x) dx separately and then we can apply those values in the first equation.

∫ e^2x (cos 3x) dx

u = cos 3 x                 dv = e^2x

du = -3 sin 3x             v = e^2x/2

= (cos 3 x)(e^2x/2) - ∫(e^2x/2)(-3 sin 3x) dx

= (cos 3 x)(e^2x/2) + (3/2) ∫e^2x sin 3x dx

so far we got the integration value of ∫ e^2x (cos 3x) d,now we are going to plug this value in the first equation

= (sin 3 x)(e^2x/2)-(3/2)[(cos 3 x)(e^2x/2) + (3/2) ∫e^2x sin 3x dx]

= (sin 3 x)(e^2x/2)-(3/4)(cos 3 x)(e^2x) - (9/4) ∫e^2x sin 3x dx

[1 + (9/4)]∫e^2x sin 3x dx = (sin 3 x)(e^2x/2)-(3/4)(cos 3 x)(e^2x) + C

(13/4) ∫e^2x sin 3x dx = (sin 3 x)(e^2x/2)-(3/4)(cos 3 x)(e^2x)

(13/4) ∫e^2x sin 3x dx = (e^2x/2)[sin 3 x-(3/2)(cos 3 x)]

(13/4) ∫e^2x sin 3x dx = (e^2x/2)[2sin 3 x-3cos 3 x]/2

(13/4) ∫e^2x sin 3x dx = (e^2x/4)[2sin 3 x-3cos 3 x]

∫e^2x sin 3x dx = (4/13)(e^2x/4)[2sin 3 x-3cos 3 x]

∫e^2x sin 3x dx = (e^2x/13)[2sin 3 x-3cos 3 x] + C

Question 19

Integrate the following with respect to x,   e^x cos 2 x

Solution:

∫ e^x cos 2x dx

u = cos 2x              dv = e^x

du = -2 sin 2 x        v = e^x

∫ u dv = u v - ∫ v du

= (cos 2x)(e^x) - ∫(e^x) (-2 sin 2 x) dx

= e^x(cos 2x) + 2∫ e^x (sin 2x) dx--------(1)

now we are going to find the integration value of ∫ e^x (sin 2x) dx separately and then we can apply those values in the first equation.

∫ e^x (sin 2x) dx

u = sin 2x                 dv = e^x

du = 2 cos 2x             v = e^x

= (sin 2x)(e^x) - ∫(e^x)(2 cos 2x) dx

= (sin 2x)(e^x) - 2 ∫(e^x)(cos 2x) dx

now we are going to apply this value in the first equation

= e^x(cos 2x) + 2[(sin 2x)(e^x) - 2 ∫(e^x)(cos 2x) dx]

∫ e^x cos 2x dx = e^x(cos 2x) + 2 e^x sin 2x - 4 ∫e^x cos 2x dx]

∫ e^x cos 2x dx + 4 ∫e^x cos 2x dx = e^x(cos 2x) + 2 e^x sin 2x

5 ∫ e^x cos 2x dx = e^x[cos 2x + 2 sin 2x]

e^x cos 2x dx = (e^x[/5) [cos 2x + 2 sin 2x]

integration worksheet6 solution8 integration worksheet6 solution8