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