INTEGRAL OF E POWER AX Sin bx OR Cos bx

Example 1 :

Integrate 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 3x) dx

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

∫ e^2x (cos 3x) dx

u  =  cos 3 x                 dv  =  e^2x

du  =  -3sin 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

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

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

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

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

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

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

Example 2 :

Integrate e^x cos 2x

Solution :

∫ e^x cos 2x dx

u  =  cos 2x              dv  =  e^x

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

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