Integration by parts formula

3 minutes, 19 seconds Read

Integration by parts formula

Integration by parts formula


If u and vare two function of x then

∫(uv)dx =[u.∫vdx]-∫{du/dx.∫vdx}dx


For any two functions f₁(x)andf₂(x),we have





Let f₁(x)=u and f₂'(x)=v so that f₂(x)=∫vdx


We can express this result as given below:

Integral of product of two functions

= (1st function ) X (integral of 2nd)- ∫{(derivative of 1st) x (integral of 2nd)} dx


(i) If the integrand is of the form f(x)xⁿ, we consider xⁿ as the first function and f(x) as the second function.

(ii) If the integrand contains a logarithmic or an inverse trigonometric function, we take it as the first function. In all such cases, if the second function is not given, we take it as 1.

For e.g

(i) ∫xsec²dx

solution Integration by parts,taking x as the first function,we get



             =xtanx+log |cosx|+C

∫xsec²dx =xtanx+log |cosx|+C

(ii) ∫x²sinxdx

Solution Integration by parts,taking x² as the first function,we get

∫x²sinxdx= x∫sinxdx-∫{d/dx(x²).∫sinxdx}dx




(Integration x(cosx) by parts)


∫x²sinxdx=-x²(cosx)+2[x sinx+cosx] +C

(iii) ∫e²ˣsinxdx

solution Integration by parts, we get





= – e²ˣ (cosx) + 2 [ (e²ˣ∫(cosx) dx) -∫{d/dx(e²ˣ).∫(cosx) dx} dx ]

                   (Integration  e²ˣ(cosx) by parts)


∴ 5∫e²ˣsinxdx= e²ˣ(2sinx-cosx)+C

   ∫e²ˣsinxdx= (1/5)e²ˣ(2sinx-cosx)+C

Integrals of the form ∫eˣ[f(x)+f'(x)]dx


∫eˣ[f(x)+f'(x)]dx =eˣ.f(x) +C

∫eˣ[f(x)+f'(x)]dx =∫eˣ.f(x)dx +∫eˣf'(x)dx

=f(x).∫eˣdx -∫{f'(x).∫eˣdx}dx+∫eˣf'(x)dx +C

(evaluating the first integral by parts)

=f(x)eˣ -∫eˣf'(x)dx+∫eˣf'(x)dx +C

= eˣf(x) +C

∫eˣ[f(x)+f'(x)]dx =eˣ.f(x) +C

for e.g

(iv) ∫eˣ(tanx+logsecx)dx




          (  where f(x)=log(secx) and

                    f'(x)=(1/secx)secx.tanx =tanx)

          =eˣ.f(x) +C 

         = eˣlog(secx)+C

∴ ∫eˣ(tanx+logsecx)dx = eˣlog(secx)+C

Integrals of the form

eᵏˣ.{k.f(x)+f'(x)}dx =eᵏˣ. f(x)


Proof ∫eᵏˣ.{k.f(x)+f'(x)}dx

                            = k.∫eᵏˣf(x)dx+∫eᵏˣf'(x)dx

=k.[eᵏˣ/k. f(x)-∫f'(x).eᵏˣ/k dx]+∫eᵏˣ f'(x)dx+C

 (evaluating the first integral by parts)

=eᵏˣ. f(x)-∫eᵏˣ. f'(x)dx+∫eᵏˣ f'(x)dx+C

=eᵏˣ. f(x) +C

  ∴ ∫eᵏˣ.{k.f(x)+f'(x)}dx =eᵏˣ. f(x)

for e.g

(v) ∫e²ˣ.{2cosx-sinx}dx

Solution we have


  =2.∫cosx.e²ˣ dx-∫sinx.e²ˣdx

 =2.[cosx.e²ˣ/2 -∫-sinx.e²ˣ/2dx]-∫e²ˣsinx dx

 ( integrating e²ˣcosx by parts)

 =cosx.e²ˣ+∫sinx.e²ˣdx]-∫e²ˣsinx dx+C


∴ ∫e²ˣ.{2cosx-sinx}dx =cosx.e²ˣ+C

integration by parts formula

ये भी पढ़े….

integration by parts formula

Similar Posts

Leave a Reply