Integration by Parts: Derivation, Application and Questions

Integration by parts refers to a unique integration method, which is extremely useful when multiplying two functions together. It is used whenever you have an integral of a product and, in that case, direct antidifferentiation has failed. The product rule for differentiation in reverse is used here. This method is beneficial for finding integrals by reducing the terms into their standard forms. For instance, to find the integration of x sin x, the integration method is used. Knowing just the formula is not enough, entrance exams like JEE Main and NEET require conceptual knowledge based on which questions are asked in the exam. The utility of this integration method is beneficial in many ways, as well. The rule of this method is:

∫u v dx = u∫v dx −∫u' (∫v dx) dx.

Here, 
u = the function u (x),
v = the function v (x),
u' =  the derivative of function u (x).

Exams like CUET and IISER often ask questions which are application based. There are no direct questions on formula or derivation. This makes understanding all steps really important. Let us now derive the formula of integration by parts using the product rule:

Integration by parts typically uses the Ilate Rule, which helps to find the first and second functions in the same method. In integration by parts, when the product of two functions is given, apply the particular formula. The integral of these two functions is taken by considering them as the first and second functions. This arrangement is known as the Ilate Rule. 

The first integration by parts begins with
d (u v) = u d v + v d u.

Integrating on both the sides,
∫ d (u v) = u v = ∫ u d v + ∫ v d u.

Rearranging the above equation gives
∫ u d v = u v - ∫ v d u.

Definite integral

An integral having upper and lower limits is known as a definite integral. On a real plane, x is limited to lie. The other name of the definite integral is Riemann integral.

Indefinite integral

The indefinite integral is defined without any upper or lower limits, and they are represented as- ∫f(x)dx = F(x) + C

Where, C = constant

Function f(x) is integrated

Let us understand the applications of integration by parts:

• Integration by parts turns a complex product integral into two easier solvable integers: $\int udv=uv-\int vdu$
• This method helps in handling logarithms / inverse-trig integrals (example shown for ∫ ln x dx): $\int \ln xdx=x\ln x-x$
• It also helps in building reduction formulas (sine-power example): $I_n=\frac{n-1}{n}{I}_{n-2}$
• Integration by parts help in solving cyclic integrals such as ∫ e^x sin x dx: $\int e^x\sin xdx=\frac{1}{2}{e}^x(\sin x-\cos x)$
• This method is useful in evaluating definite integrals where boundary terms vanish (γ = 1! example): $\underset{0}{\overset{\infty }{\int }}x{e}^{-x}dx=1$
• Integration by parts helps in finding moments/expectations in probability : $E\left[X\right]=\underset{0}{\overset{\infty }{\int }}(1-F(x\left)\right)dx$
• One can derive Laplace/Fourier transform identities: $\underset{0}{\overset{\infty }{\int }}{f}^{\prime }\left(t\right)e^{-st}dt=-f\left(0\right)+s\underset{0}{\overset{\infty }{\int }}f\left(t\right)e^{-st}dt$
• It is used to prove orthogonality in Sturm–Liouville problems: $\underset{a}{\overset{b}{\int }}p\left(x\right)y_m\left(x\right)y_n\left(x\right)dx=0,m\ne n$
• Beta and Gamma functions can be linked with integration by parts: $B(m,n)=(m-1)B(m-1,n)$
• It justifies energy conservation in the 1-D wave equation: $E\left(t\right)=\frac{1}{2}\underset{-}{\overset{\infty }{\int }}{u_t}^2+c^2{u_x}^{2}dx$

 

1. Integrate  ∫ x sinX.dx

Solution. Let,

u = x, dv= sinX.dx

du = (1) dx, and v= - cosX

∫ x sinX.dx = x (-cosX) - ∫ (- cosX) dx

= - x cosX + ∫ CosX.dx

= - x cosX + sinX + C

2. Integrate  ∫ x In x.dx.

Solution. Let,

u= In x, and dv= x.dx

du = ½ dx, and v= x2/2

3. Find ∫ x cosX

Solution. Let,

x = u, after differentiation, du/dx = 1

 = x and dv/dx = cosX, already found the value i.e. - du/dx = 1

Now, since dv/dx = cosX

Integrating both sides,

v = ∫ cosx.dx

v = sinX

Formula- ∫u (dv/dx) dx = uv – ∫v (du/dx)dx

∫x cosX.dx = x sinX – ∫ sinX.1 dx 

∫x cosX.dx = x sinX + cosX + c,

where c = constant.