Mathematics Integration By Parts
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Class 12 Chapter 7 - Integrals

Integration By Parts

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Integration by Parts

● If `u` and `v` are any two differentiable functions of a single variable `x` (say). Then, by the product rule of differentiation, we have

`color{green}{d/(dx) (uv) = u (dv)/(dx) + v (du)/(dx)}`

`=>` Integrating both sides, we get

` uv = ∫ u (dv)/(dx) dx + ∫ v (du)/(dx) dx`

`=>` ` color{red} {∫ u (dv)/(dx) dx = uv - ∫ v (du)/(dx) dx}` ........(1)

Let ` u = f(x)` and ` (dv)/(dx) = g(x)`. Then

` (du)/(dx) = f'(x)` and ` v = ∫ g(x) dx`

Therefore, expression (1) can be rewritten as

`color{blue}{ ∫ f (x) g(x) dx = f (x)∫ g(x) dx – ∫[{∫ g(x) dx }f ′(x) dx]}`


i.e.,`color{blue}{ ∫ f (x) g (x) dx = f (x)∫ g (x) dx – ∫[ f ′ (x) ∫ g(x) dx] dx}`

Remember : If we take `f` as the first function and `g` as the second function, then this formula may be stated as follows:

`color{orange}{"The integral of the product of two functions = (first function) × (integral of the second function) "}`
`color{orange}{"– Integral of [(differential coefficient of the first function) × (integral of the second function)]"}`
Q 3115778669

Find ` int x cos x dx`

Class 12 Chapter 7 Example 17
Solution:

Put f (x) = x (first function) and g (x) = cos x (second function).
Then, integration by parts gives

` ∫ x cos x dx = x ∫ cos x dx = ∫ [ d/(dx) (x) ∫ cos x dx ] dx`

` = x sin x – ∫sin x dx = x sin x + cos x + C`
Suppose, we take f (x) = cos x and g(x) = x. Then

`∫ x cos x dx = cos x ∫ x dx - ∫ [ d/(dx) (cos x ) ∫ x dx ] dx`

` = (cos x) x^2/2 + ∫ sin x x^2/2 dx`

Thus, it shows that the integral ∫ x cos x dx is reduced to the comparatively more
complicated integral having more power of x. Therefore, the proper choice of the first
function and the second function is significant.
Q 3125878761

Find ` ∫ log x dx`
Class 12 Chapter 7 Example 18
Solution:

To start with, we are unable to guess a function whose derivative is log x. We
take log x as the first function and the constant function 1 as the second function. Then,
the integral of the second function is x.

Hence,` ∫ (log x * 1) dx = log x ∫ 1 dx - ∫ [ d/(dx) (log x) ∫ 1 dx ] dx`

` = (log x) * x - ∫ 1/x x dx = x log x - x +C`
Q 3135878762

Find ` ∫ x e^x dx`
Class 12 Chapter 7 Example 19
Solution:

Take first function as x and second function as `e^x`. The integral of the second
function is `e^x`.

Therefore, ` ∫ x e^x dx = x e^x - ∫ 1 * e^x dx = x e^x - e^x +C`
Q 3155878764

Find ` ∫ (x sin^(-1) x )/(sqrt (1-x^2 ) ) dx`
Class 12 Chapter 7 Example 20
Solution:

Let first function be `sin^(– 1) x` and second function be `x/(sqrt (1-x^2 ) )`

First we find the integral of the second function, i.e., ` ∫ (x dx)/( sqrt (1-x^2 ) )`

Put `t =1 – x^2`. Then dt = – 2x dx

Therefore, ` ∫ (x dx)/( sqrt (1-x^2 ) ) = -1/2 ∫ (dt)/(sqrt t) = - sqrt t = - sqrt (1-x^2 )`

Hence, ` ∫ (x sin^(-1) x)/(sqrt (1-x^2) ) dx = (sin^(-1) x) ( - sqrt (1-x^2) ) -∫ 1/(sqrt (1-x^2) ) (- sqrt (1-x^2) ) dx`

`= - sqrt (1-x^2) sin^(-1) x+x+ C = x- sqrt (1-x^2) sin^(-1) x +C`

Alternatively, this integral can also be worked out by making substitution `sin^(–1 ) x = θ` and
then integrating by parts.
Q 3175878766

Find ` ∫ e^x sin x dx`
Class 12 Chapter 7 Example 21
Solution:

Take `e^x` as the first function and sin x as second function. Then, integrating
by parts, we have

`I= ∫ e^x sin x dx = e^x (- cos x) + ∫ e^x cos x dx`

`= - e^x cos x +I_1 ` (say ) .....(1)

Taking `e^x` and cos x as the first and second functions, respectively, in `I_1`, we get

`I_1 = e^x sin x – ∫ e^x sin x dx`

Substituting the value of `I_1` in (1), we get

`I = – e^x cos x + e^x sin x – I` or `2I = e^x (sin x – cos x)`

Hence, `I = ∫ e^x sin x dx = e^x/2 (sin x - cos x ) +C`

Alternatively, above integral can also be determined by taking sin x as the first function
and `e^x` the second function.
Q 3115078860

Find

(i) ` ∫ e^x (tan^(-1) x +1/(1+x^2) ) dx`

(ii) `∫ ( (x^2 +1 ) e^x)/( (x+1)^2 ) dx`
Class 12 Chapter 7 Example 22
Solution:

(i) We have `I = ∫ e^x (tan^(-1) x + 1/(1+x^2) ) dx`

Consider `f (x) = tan^(– 1) x`, then `f ′(x) = 1/(1+ x^2 )`

Thus, the given integrand is of the form `e^x [ f (x) + f ′(x)]`.

Therefore, `I = ∫ e^x (tan^(-1) x +1/(1+x^2) ) dx = e^x tan^(-1) x +C`

(ii) We have `I = ∫ ( x^2+1 )/( (x+1)^2 ) dx = ∫ e^x [ (x^2 -1+1+1)/( (x+1)^2 ) ] dx`

`= ∫ e^x [ ( x^2 -1 )/( (x+1 )^2 ) +2/( (x+1)^2 ) ] dx = ∫ e^x [ (x-1)/(x+1) +2/(x+1)^2] dx`

Consider `f (x ) = (x-1)/(x+1)`, then ` f ' (x) = 2/( (x+1)^2 )`

Thus, the given integrand is of the form `e^x [f (x) + f ′(x)]`.

Therefore, ` ∫ (x^2 +1 )/( (x+1)^2 ) e^x dx = (x-1)/(x+1) e^x +C`
 
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