Integration by Parts - Formula, Derivation, Applications, Examples

Integration By Parts:

Integration is the reverse process of differentiation. In integration, we find the function whose differential coefficient is given. The rate of change of a quantity y concerning another quantity x is called the derivative or differential coefficient of y concerning x. Geometrically, the Differentiation of a function at a point represents the slope of the tangent to the graph of the function at that point.

For example,

$\begin{aligned} & \frac{d}{d x}(\sin x)=\cos x \\ & \frac{d}{d x}\left(x^2\right)=2 x \\ & \frac{d}{d x}\left(e^x\right)=e^x\end{aligned}$

To evaluate the integration of a function which is a product of two functions, we use the method of integration by parts.

If two functions of x, u, and v are given,

where u is function of u(x),

v is function of v(x)

then

$\int u v d x=u \times \int v d x-\int\left\{\frac{d u}{d x} \int v d x\right\} d x$

i.e.

The integral of the product of two functions =

( first function) X ( integral of second function ) - integral of (differential of first function X integral of second function).

Proof:

Let, h(x) = f(x) . g(x), then by using the product rule, we obtain, h′(x) = f ′(x) g(x) + g′(x) f(x). Let’s now integrate both sides of this equation:

$\int h^{\prime}(x) d x=\int\left(g(x) f^{\prime}(x)+f(x) g^{\prime}(x)\right) d x$

this gives us,

$h(x)=f(x) g(x)=\int g(x) f^{\prime}(x) d x+\int f(x) g^{\prime}(x) d x$

Now we solve for $\int f(x) g^{\prime}(x) d x$

$\int f(x) g^{\prime}(x) d x=f(x) g(x)-\int g(x) f^{\prime}(x) d x$

By making the substitutions $u=f(x)$ and $v=g^{\prime}(x)$ which in turn make $g(x)=\int v d x$, we have the more compact form

$\int u v d x=u \int v d x-\int\left\{\frac{d u}{d x} \int v d x\right\} d x$

Important point for selecting first function (u) and second function (v)

Usually, we use the following preference order for selecting the first function. (Inverse, Logarithmic, Algebraic, Trigonometric, Exponential). In the above-stated order, the function on the left is always chosen as the first function. This rule is called ILATE.

For example, if an integral contains a logarithmic function and an algebraic function, we should choose the logarithmic function as the first function (u).

One of the interesting application of integration by part is:

$\int \mathrm{e}^{\mathbf{x}}\left\{\mathbf{f}(\mathbf{x})+\mathbf{f}^{\prime}(\mathbf{x})\right\} \mathrm{dx}=\mathrm{e}^{\mathbf{x}} \mathbf{f}(\mathbf{x})+\mathbf{C}$

Proof:
We have given $\int \mathrm{e}^x\left\{f(x)+f^{\prime}(x)\right\} d x$

$=\int \mathrm{e}^{\mathrm{x}} \cdot \mathrm{f}(\mathrm{x}) \mathrm{dx}+\int \mathrm{e}^{\mathrm{x}} \cdot \mathrm{f}^{\prime}(\mathrm{x}) \mathrm{dx}$

In first integral, take $f(x)$ as first function and $e^x$ as second function

$\begin{aligned}
& =f(x) \cdot e^x-\int f^{\prime}(x) \cdot e^x d x+\int e^x \cdot f^{\prime}(x) d x+C \\
& =f(x) \cdot e^x+C
\end{aligned}$

Thus, to evaluate the integrals of the type $\int e^x\left\{f(x)+f^{\prime}(x)\right\} d x$ we first express the integral as the sum of two integrals $\int e^x f(x) d x$ and $\int e^x f^{\prime}(x) d x$ and then integrate the integral involving $e^x f(x)$ as integral by parts taking $e^x$ as second function.

General Formula

$\int \mathrm{e}^{g(x)} \cdot\left\{f(x) g^{\prime}(x)+f^{\prime}(x)\right\} d x=f(x) \cdot e^{g(x)}+C$

Integration of $\int e^{a x} \cdot \sin (b x) d x$ and $\int e^{a x} \cdot \cos (b x) d x$
Let, $\quad \mathrm{I}=\int \mathrm{e}^{a x}(\sin b x) \mathrm{dx}$
take $\sin b x$ as first function and $\mathrm{e}^{a x}$ as second

$\begin{aligned}
& =\sin b x \cdot\left(\frac{e^{a x}}{a}\right)-\int b \cos b x \cdot \frac{e^{a x}}{a} d x \\
& =\frac{1}{a} \sin b x \cdot e^{a x}-\frac{b}{a}\left\{\cos b x \cdot \frac{e^{a x}}{a}-\int(-b \sin b x) \cdot \frac{e^{a x}}{a} d x\right\} \\
& =\frac{1}{a} \sin b x \cdot e^{a x}-\frac{b}{a^2} \cos b x \cdot e^{a x}-\frac{b^2}{a^2} \int \sin b x \cdot e^{a x} d x \\
I & =\frac{1}{a} \sin b x \cdot e^{a x}-\frac{b}{a^2} \cos b x \cdot e^{a x}-\frac{b^2}{a^2} I \\
\therefore \quad I+\frac{b^2}{a^2} I & =\frac{1 \cdot e^{a x}}{a^2} \cdot(a \sin b x-b \cos b x) \\
\Rightarrow \quad I & =\frac{e^{a x}}{a^2+b^2}(a \sin b x-b \cos b x)+C
\end{aligned}$

In the same way,

$\int e^{a x} \cos b x d x=\frac{e^{a x}}{a^2+b^2}(a \cos b x+b \sin b x)+C$
Result for integration by parts or ex :

$\begin{aligned}
& \int e^{\tan ^{-1} x}\left[\frac{x^n+1}{x^2+1}+n x^{n-1}\right] d x=e^{\tan ^{-1} x}\left(x^n+1\right)+c \\
& \int e^{g(x)}\left\{f(x) g^{\prime}(x)+f^{\prime}(x)\right\} d x=f(x) \cdot e^{g(x)}+C
\end{aligned}$
developments in many scientific and engineering disciplines.

Recommended Video Based on Integration by Parts

Solved Examples Based On Integration by Parts:

Example 1: The integral $\int x \cos ^{-1}\left(\frac{1-x^2}{1+x^2}\right) d x(x>0)$ is equal to:
${ }_1-x+\left(1+x^2\right) \tan ^{-1} x+c$
2) $x-\left(1+x^2\right) \cot ^{-1} x+c$
3) $-x+\left(1+x^2\right) \cot ^{-1} x+c$
4) $x-\left(1+x^2\right) \tan ^{-1} x+c$

Solution

As learned in the concept

Integration By PARTS -

Let $u$ and $v$ are two functions then

$
\int u \cdot v d x=u \int v d x-\int\left(\frac{d u}{d x} \int v d x\right) d x
$

- wherein

Where $u$ is the lst function $v$ is the Ind function

$
\begin{aligned}
& \text { Put } x=\tan \theta ; d x=\sec ^2 \theta d \theta \\
& \int x \cos ^{-1}\left(\frac{1-x^2}{1+x^2}\right) d x(x>0) \\
& \int \tan \theta\left[\cos ^{-1}\left(\frac{1-\tan ^2 \theta}{1+\tan ^2 \theta}\right)\right]\left(\sec ^2 \theta d \theta\right) \\
& \int \tan \theta\left[\cos ^{-1}(\cos 2 \theta)\right]\left(\sec ^2 \theta d \theta\right) \\
& \int \tan \theta \cdot 2 \theta \sec ^2 \theta d \theta=2 \int \theta\left(\tan \theta \sec ^2 \theta\right) d \Theta
\end{aligned}
$

Put $\tan \theta=t$ we get $\sec ^2 \theta d \theta=d t$

$
\begin{aligned}
& 2 \int \tan ^{-1} t(t) d t \\
& 2 \int t \cdot \tan ^{-1} t \cdot d t
\end{aligned}
$

Using by parts

$\begin{aligned}
& 2\left[\frac{t^2}{2} \cdot \tan ^{-1} t-\int \frac{1}{1+t^2} \cdot \frac{t^2}{2} \cdot d t\right] \\
& t^2 \cdot \tan ^{-1} t-\int \frac{1}{1+t^2} \cdot t^2 \cdot d t \\
& t^2 \cdot \tan ^{-1} t-\int \frac{t^2+1-1}{1+t^2} \cdot d t \\
& t^2 \cdot \tan ^{-1} t-\int\left[1-\frac{1}{1+t^2}\right] \cdot d t \\
& t^2 \cdot \tan ^{-1} t-\left[t-\tan ^{-1} t\right] \\
& \Rightarrow x=t \\
& x=\tan \theta \quad \& \quad \tan \theta=t \\
& \int x \cos ^{-1}\left(\frac{1-x^2}{1+x^2}\right) d x=x^2 \tan ^{-1} x-x+\tan ^{-1} x+C
\end{aligned}$
Hence, the answer is the option 1.

Example 2: If $\int f(x) d x=\Psi(x)$, then $\int x^5 f\left(x^3\right) d x$ is equal to:
1) $\frac{1}{3}\left[x^3 \Psi\left(x^3\right)-\int x^3 \Psi\left(x^3\right) d x\right]+C$
2) $\frac{1}{3}\left[x^3 \Psi\left(x^3\right)-\int x^2 \Psi\left(x^3\right) d x\right]+C$
3) $\frac{1}{3} x^3 \Psi\left(x^3\right)-3 \int x^3 \Psi\left(x^3\right) d x+C$
4) $\frac{1}{3} x^3 \Psi\left(x^3\right)-\int x^2 \Psi\left(x^3\right) d x+C$

Solution
Using Integration By Parts,

$\begin{aligned}
& \int f(x) d x=\Psi(x) \int x^5 f\left(x^3\right) d x=\frac{1}{3} \int\left(3 x^2\right) \cdot x^3 f\left(x^3\right) d x P u t x^3=t ; 3 x^2 d x=d t \\
& \frac{1}{3} \int t f(t) d t=\frac{1}{3}\left[t \int f(t) d t-\int 1 \int f(t) d t\right]
\end{aligned}$

Replace $t$ by $x^3$

$I=\frac{1}{3} x^3 \Psi\left(x^3\right)-\frac{1}{3} \int \Psi\left(x^3\right) d t=\frac{1}{3} x^3 \Psi\left(x^3\right)-\frac{1}{3} \int 3 x^2 d x . \Psi\left(x^3\right)=\left[\frac{1}{3} x^3 \Psi\left(x^3\right)-\int x^2 \Psi\left(x^3\right) d x\right]+c$
Hence, the answer is the option (4).

Example 3: Let $f(x)=\int \frac{\sqrt{x}}{(1+x)^2} d x(x \geqslant 0)$.Then $f(3)-f(1)$ is equal to:
1) $-\frac{\pi}{12}+\frac{1}{2}+\frac{\sqrt{3}}{4}$
2) $\frac{\pi}{6}+\frac{1}{2}+\frac{\sqrt{3}}{4}$
3) $-\frac{\pi}{6}+\frac{1}{2}+\frac{\sqrt{3}}{4}$
4) $\frac{\pi}{12}+\frac{1}{2}-\frac{\sqrt{3}}{4}$

Solution
$f(x)=\int_1^3 \frac{\sqrt{x} d x}{(1+x)^2}=\int_1^{\sqrt{3}} \frac{t \cdot 2 t d t}{\left(1+t^2\right)^2}($ put $\sqrt{x}=t)$
$=\left(-\frac{t}{1+t^2}\right)_t^{\sqrt{3}}+\left(\tan ^{-1} t\right)_1^{\sqrt{3}}$ [Applying by parts ]
$=-\left(\frac{\sqrt{3}}{4}-\frac{1}{2}\right)+\frac{\pi}{3}-\frac{\pi}{4}$
$=\frac{1}{2}-\frac{\sqrt{3}}{4}+\frac{\pi}{12}$

Hence, the answer is the option 4.

Example 4: The integral $\int\left(\frac{x}{x \sin +\cos x}\right)^2 d x$ is equal to (where C is constant of integration).
1) $\tan x-\frac{x \sec x}{x \sin x+\cos x}+C$
2) $\sec x \frac{x \tan x}{x \sin x+\cos x}+C$
3) $\sec x-\frac{x \tan x}{x \sin x+\cos x}+C$
4) $\tan x+\frac{x \sec x}{x \sin x+\cos x}+C$

Solution

$\begin{aligned}
& \int\left(\frac{x}{x \sin x+\cos x}\right)^2 d x=\int\left(\frac{x}{\cos x}\right) \cdot \frac{x \cos x d x}{(x \sin x+\cos x)^2} \\
& =\frac{x}{\cos x}\left(-\frac{1}{x \sin x+\cos x}\right)+\int\left(\frac{\cos x+x \sin x}{\cos ^2 x}\right)\left(\frac{1}{x \sin x+\cos x}\right) d x \\
& =-\frac{x \sec x}{x \sin x+\cos x}+\int \sec ^2 x d x \\
& =-\frac{x \sec x}{x \sin x+\cos x}+\tan x+C
\end{aligned}$
Hence, the answer is option (1).

Example 5 :

$\int_{\mathrm{If}}^1\left(e^{2 x}+2 e^x-e^{-x}-1\right) e^{\left(e^x+e^{-x}\right)} d x=g(x) e^{\left(e^x+e^{-x}\right)}+c$

1) e
2) $e^2$
3) 1
4) 2

Solution:

$\begin{aligned}
& e^{2 x}+2 e^x-e^{-x}-1 \\
& =e^x\left(e^x+1\right)-e^{-x}\left(e^x+1\right)+e^x \\
& \left.=\left[\left(e^x+1\right)\right]\left(e^x-e^{-x}\right)+e^x\right] \\
& \text { so } I=\int\left(e^x+1\right)\left(e^x-e^{-x}\right) e^{e^x+e^{-x}}+\int e^x \cdot e^{e^x+e^{-x}} d x \\
& =\left(e^x+1\right) e^{e^x+e^{-x}}-\int e^x \cdot e^{e^x+e^{-x}} d x+\int e^x \cdot e^{e^x+e^{-x}} d x \\
& =\left(e^x+1\right) e^{e^x+e^{-x}}+C \\
& \therefore g(x)=e^x+1 \Rightarrow g(0)=2
\end{aligned}$

Hence, the answer is option (4).