Integration By Parts Formula - Practice Questions & MCQ

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, 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 $\mathrm{u}=\mathrm{f}(\mathrm{x})$ and $\mathrm{v}=\mathrm{g}^{\prime}(\mathrm{x})$ which in turn make $g(x)=\int v d x$, we have the more compact form
$
\int \mathrm{uvdx}=\mathrm{u} \int \mathrm{vdx}-\int\left\{\frac{\mathrm{du}}{\mathrm{dx}} \int \mathrm{vdx}\right\} \mathrm{dx}
$

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 as 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 applications of integration by part is:

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

Proof:

We have given $\int \mathrm{e}^{\mathrm{x}}\left\{\mathrm{f}(\mathrm{x})+\mathrm{f}^{\prime}(\mathrm{x})\right\} \mathrm{dx}$
$
=\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 $\mathrm{f}(\mathrm{x})$ as first function and $\mathrm{e}^{\mathrm{x}}$ as second function
$
\begin{aligned}
& =\mathrm{f}(\mathrm{x}) \cdot \mathrm{e}^{\mathrm{x}}-\int \mathrm{f}^{\prime}(\mathrm{x}) \cdot \mathrm{e}^{\mathrm{x}} \mathrm{dx}+\int \mathrm{e}^{\mathrm{x}} \cdot \mathrm{f}^{\prime}(\mathrm{x}) \mathrm{dx}+C \\
& =\mathrm{f}(\mathrm{x}) \cdot \mathrm{e}^{\mathrm{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 a 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, $\quad I=\int e^{a x}(\sin b x) d x$
take $\sin b x$ as first function and $\mathrm{e}^{\mathrm{ax}}$ 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 \mathrm{I}+\frac{\mathrm{b}^2}{\mathrm{a}^2} \mathrm{I}=\frac{1 \cdot \mathrm{e}^{\mathrm{ax}}}{\mathrm{a}^2} \cdot(\mathrm{a} \sin \mathrm{bx}-\mathrm{b} \cos \mathrm{bx}) \\
& \Rightarrow \quad \mathrm{I}=\frac{\mathrm{e}^{\mathrm{ax}}}{\mathrm{a}^2+\mathrm{b}^2}(\mathrm{a} \sin \mathrm{bx}-\mathrm{b} \cos \mathrm{bx})+\mathrm{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
$