Integrals of the form e pow(ax) cos bx and e pow(ax) sin bx

Integrals of the form

(i) ∫e ax ax sinbx dx (ii) ∫e ax cosbx dx

The following examples illustrate that there are some integrals whose integration continues forever. Whenever we integrate function of the form e axcos bx or eax sinbx, we have to apply the Integration by Parts rule twice to get the similar integral on both sides to solve.

Result 11.1

Caution

In applying integration by parts to specific integrals, the pair of choice for u and dv once initially assumed should be maintained for the successive integrals on the right hand side. (See the above two examples). The pair of choice should not be interchanged.

Examples 11.36

Evaluate the following integrals

(i) ∫e 3x cos 2x dx

(ii) ∫e −5x sin 3x dx

(i) ∫e 3x cos 2x dx

Using the formula

EXERCISE 11.8

Integrate the following with respect to x

(1) (i) eax cos bx (ii) e2 x sin x (iii) e− x cos 2x

(2)  (i) e−3x sin 2x (ii) e−4xsin 2x (iii) e−3x cos x

Result 11.2

∫ex [ f (x) + f ′(x)]dx = ex f (x) + c

Proof

Let I = ∫ex [ f (x) + f ′(x)]dx

= ∫ e x f (x)dx + ∫ex f ′(x)dx

Take u = f ( x); du =  f’ ( x) dx, in the first integral

dv = ex ;v = ex ,

That is, I = ex f (x) − ∫ e x f ′( x)dx + ∫ex f ′(x)dx + c

Therefore, I = e x f ( x ) + c.

EXERCISE 11.9

Integrate the following with respect to x: