3.0 Integration by parts

  Indefinite Integrals

• The integral $\int {vdx} $ will not contain any arbitrary constant.
• If in the product of two functions, one function is not directly integrable, then consider that function as the first function $u$ and remaining function as second function $v$. For example: In $\int {x{{\sin }^{ - 1}}xdx} $, ${\sin ^{ - 1}}x$ is considered as first function and $x$ as the second function.
• If in the product there is only one function which is not directly integrable, then consider other function to be $1$.
• The choice of $f(x)$ and $g(x)$ is decided by ILATE rule. The following preference order is considered for selecting the first function $f(x)$: $$\begin{equation} \begin{aligned} I \to {\text{Inverse function}} \\ L \to {\text{Logarithmic function}} \\ A \to {\text{Algebraic function}} \\ T \to {\text{Trigonometric function}} \\ E \to {\text{Exponential function}} \\\end{aligned} \end{equation} $$

Important result

$\int {{e^{ax}}\sin bxdx} $

$$\begin{equation} \begin{aligned} I = \int {{e^{ax}}(\sin bx)dx} \\ I = \int {\sin bx.{e^{ax}}} dx \\\end{aligned} \end{equation} $$

Apply integration by parts, we get

$$\begin{equation} \begin{aligned} I = \sin bx.\left( {\frac{{{e^{ax}}}}{a}} \right) - \int {b\cos bx.} \frac{{{e^{ax}}}}{a}dx \\ I = \frac{1}{a}\sin bx.{e^{ax}} - \frac{b}{a}\left\{ {\cos bx.\frac{{{e^{ax}}}}{a} - \int {( - b\sin bx).\frac{{{e^{ax}}}}{a}dx} } \right\} \\ I = \frac{1}{a}\sin bx.{e^{ax}} - \frac{b}{{{a^2}}}\cos bx.{e^{ax}} - \frac{{{b^2}}}{{{a^2}}}\int {\sin bx.{e^{ax}}} dx \\ I = \frac{1}{a}\sin bx.{e^{ax}} - \frac{b}{{{a^2}}}\cos bx.{e^{ax}} - \frac{{{b^2}}}{{{a^2}}}I \\ I + \frac{{{b^2}}}{{{a^2}}}I = \frac{{{e^{ax}}}}{{{a^2}}}\left( {a\sin bx - b\cos bx} \right) \\ I = \frac{{{e^{ax}}}}{{{a^2} + {b^2}}}\left( {a\sin bx - b\cos bx} \right) + C \\\end{aligned} \end{equation} $$

Similarly, $$\int {{e^{ax}}\cos bxdx} = \frac{{{e^{ax}}}}{{{a^2} + {b^2}}}\left( {a\cos bx + b\sin bx} \right) + C$$

Question 3. Evaluate

$$\int {\frac{{{{\sin }^{ - 1}}\sqrt x - {{\cos }^{ - 1}}\sqrt x }}{{{{\sin }^{ - 1}}\sqrt x + {{\cos }^{ - 1}}\sqrt x }}} dx$$

Solution:

$$\begin{equation} \begin{aligned} I = \int {\frac{{{{\sin }^{ - 1}}\sqrt x - {{\cos }^{ - 1}}\sqrt x }}{{{{\sin }^{ - 1}}\sqrt x + {{\cos }^{ - 1}}\sqrt x }}} dx \\ I = \int {\frac{{{{\sin }^{ - 1}}\sqrt x - \left( {\frac{\pi }{2} - {{\sin }^{ - 1}}\sqrt x } \right)}}{{\frac{\pi }{2}}}} dx\quad \left( {\because {{\sin }^{ - 1}}x + {{\cos }^{ - 1}}x = \frac{\pi }{2}} \right) \\ I = \frac{2}{\pi }\int {\left( {2{{\sin }^{ - 1}}\sqrt x - \frac{\pi }{2}} \right)} dx \\ I = \frac{4}{\pi }\int {{{\sin }^{ - 1}}\sqrt x } dx - \int {dx} \\ I = \frac{4}{\pi }\int {{{\sin }^{ - 1}}\sqrt x } dx - x + C \\\end{aligned} \end{equation} $$

Put $\begin{equation} \begin{aligned} x = {\sin ^2}\theta \;\; \Rightarrow dx = 2\sin \theta \cos \theta d\theta \\ \therefore dx = \sin 2\theta d\theta \\\end{aligned} \end{equation} $

$$I' = \int {{{\sin }^{ - 1}}\sqrt x } dx = \int {\theta .\sin 2\theta } d\theta $$

Now, apply integration by parts

$$\begin{equation} \begin{aligned} I' = \int {\theta .\sin 2\theta } d\theta \\ I' = \theta \int {\sin 2\theta d\theta } - \int {\left\{ {\frac{d}{{d\theta }}\left( \theta \right).\int {\sin 2\theta d\theta } } \right\}d\theta } \\ I' = - \theta \frac{{\cos 2\theta }}{2} + \frac{1}{2}\int {\cos 2\theta } d\theta \\ I' = - \frac{{\theta \cos 2\theta }}{2} + \frac{1}{4}\sin 2\theta \\ I' = - \frac{\theta }{2}\left( {1 - 2{{\sin }^2}\theta } \right) + \frac{1}{2}.\sin \theta .\cos \theta \\ I' = - \frac{\theta }{2}\left( {1 - 2{{\sin }^2}\theta } \right) + \frac{1}{2}.\sin \theta .\sqrt {1 - {{\sin }^2}\theta } \\ I' = - \frac{1}{2}{\sin ^{ - 1}}\sqrt x (1 - 2x) + \frac{1}{2}\sqrt x \sqrt {1 - x} \\\end{aligned} \end{equation} $$

Therefore, $$\begin{equation} \begin{aligned} I = \frac{4}{\pi }\left\{ { - \frac{1}{2}{{\sin }^{ - 1}}\sqrt x (1 - 2x) + \frac{1}{2}\sqrt x \sqrt {1 - x} } \right\} - x + C \\ I = \frac{2}{\pi }\left\{ {\sqrt {x - {x^2}} - (1 - 2x){{\sin }^{ - 1}}\sqrt x } \right\} - x + C \\\end{aligned} \end{equation} $$