Question

How do you find the integral \[\int {(\sin x)(\cos x)dx} \] ?

Answer 1

Hint: Here, we need to find the value of \[\int {(\sin x)(\cos x)dx} \] . There are many ways to find the answer. One should know some trigonometry formulas such as \[{\sin ^2}x + {\cos ^2}x = 1\] and \[\cos 2x = 2{\cos ^2}x - 1\] . To find integration of sin x cos x, we will use the formula \[\sin 2x = 2\sin x\cos x\] . By using this, and evaluating this, we will get the final output.

Complete step by step answer:
Given that, \[\int {(\sin x)(\cos x)dx} \]. We have many methods to solve this problem. We will use all of them one by one to solve them.The different methods used are as below:
First we will substitute \[u = \sin x\] and applying this, we will get,
Let,
\[u = \sin x\]
\[ \Rightarrow du = \cos xdx\]
For,
\[\int {(\sin x)(\cos x)dx} \]
Substitute the value of u and du in the above integral, we will get,
\[ \Rightarrow \int {(u)(du)} \]
Removing the brackets, we will get,
\[ \Rightarrow \int {udu} \]
\[ \Rightarrow \dfrac{{{u^2}}}{2} + C\] where C is the constant of integration

Again, we will substitute the value of u, we will get,
\[ \Rightarrow \dfrac{{{{(\sin x)}^2}}}{2} + C\]
\[ \Rightarrow \dfrac{{{{\sin }^2}x}}{2} + C\]
We will use the identity here as below:
\[{\sin ^2}x + {\cos ^2}x = 1\]\[ \Rightarrow {\sin ^2}x = 1 - {\cos ^2}x\]
So applying this, we will get,
\[ \Rightarrow \dfrac{{1 - {{\cos }^2}x}}{2} + C\]
\[ \Rightarrow \dfrac{1}{2} + \dfrac{{ - {{\cos }^2}x}}{2} + C\]
\[ \Rightarrow - \dfrac{{{{\cos }^2}x}}{2} + \dfrac{1}{2} + C\]
Here, \[\dfrac{1}{2}\] is absorbed into C as C represents any constant, we will get,
\[ \Rightarrow - \dfrac{{{{\cos }^2}x}}{2} + C\]

First we will substitute \[u = \sin x\] and applying this, we will get,
Let,
\[u = \cos x\]
\[ \Rightarrow du = - \sin xdx\]
For,
\[\int {(\sin x)(\cos x)dx} \]
\[ \Rightarrow\int {(\cos x)(\sin x)dx} \]
Substitute the value of u and du in the above integral, we will get,
\[ \Rightarrow \int {(u)( - du)} \]
Removing the brackets, we will get,
\[ \Rightarrow - \int {udu} \]
\[ \Rightarrow - \dfrac{{{u^2}}}{2} + C\]
Again, we will substitute the value of u, we will get,
\[ \Rightarrow - \dfrac{{{{(\cos x)}^2}}}{2} + C\]
\[ \Rightarrow - \dfrac{{{{\cos }^2}x}}{2} + C\]

Here, we will use the formula of sin2x as below:
\[\sin 2x = 2\sin x\cos x\]
\[ \Rightarrow \dfrac{{\sin 2x}}{2} = \sin x\cos x\]
\[\Rightarrow \int {(\sin x)(\cos x)dx} \]
Substitute the value, we will get,
\[\Rightarrow \int {\dfrac{{\sin 2x}}{2}dx} \]
\[ \Rightarrow \dfrac{1}{2}\int {\sin 2xdx} \]
Let, \[u = 2x\]
\[ \Rightarrow du = 2dx\]
\[ \Rightarrow dx = \dfrac{{du}}{2}\]

Now, we will substitute the values of 2x and dx, we will get,
\[ \Rightarrow \dfrac{1}{2}\int {\sin u\dfrac{{du}}{2}} \]
\[ \Rightarrow \dfrac{1}{4}\int {\sin udu} \]
We know that, \[\int {\sin xdx = - \cos x + C} \] and so applying this, we will get,
\[ \Rightarrow \dfrac{1}{4}( - \cos u) + C\]
\[ \Rightarrow - \dfrac{1}{4}\cos u + C\]
\[ \Rightarrow - \dfrac{1}{4}\cos 2x + C\]
We will use the formula \[\cos 2x = 2{\cos ^2}x - 1\] and so using this, we will get,
\[ \Rightarrow - \dfrac{1}{4}\left( {2{{\cos }^2}x - 1} \right) + C\]
Removing the brackets, we will get,
\[ \Rightarrow \dfrac{{ - {{\cos }^2}x}}{2} + \dfrac{1}{4} + C\]
On simplifying this, we will get,
\[ \Rightarrow \dfrac{{ - {{\cos }^2}x}}{2} + C\]

Hence, \[\int {(\sin x)(\cos x)dx} = \dfrac{{ - {{\cos }^2}x}}{2} + C\].

Note: Integration is a method of adding or summing up the parts to find the whole. It is a reverse process of differentiation, where we reduce the functions into parts. The integral is calculated to find the functions which will describe the area, displacement, volume, that occurs due to a collection of small data, which cannot be measured singularly. Trigonometry is a branch of mathematics that studies relationships between side lengths and angles of triangles.