Question

How do you integrate $\int{\dfrac{\sin x}{\sin x-\cos x}dx}$?

Answer 1

In the given question, we have been asked to integrate the function. In order to solve the questions, we have to use trigonometric identities to simplify the given integration function and then follow integration formulas or methods to integrate. We have to split the given trigonometric terms by using identities. Perform further integration in the simplified form and hence we get the required solution.

Formula used:
i) Following trigonometric identities are used:
${{\cos }^{2}}x-{{\sin }^{2}}x=\cos 2x$
$\sin 2x=2\ sin x\cos x$
${{\sin }^{2}}x=\dfrac{1-\cos 2x}{2}$

ii) Following integration formula are used:
$\int{\tan \left( x \right)dx}=\log \left| \sec x \right|+C$
$\int{\sec \left( x \right)dx=\log \left| \sec x+\tan x \right|}+C$
$\int{dx=x+C}$

Complete step-by-step solution:
We have the given integration function,
$\Rightarrow \int{\dfrac{\sin x}{\sin x-\cos x}dx}$
Let,
$\Rightarrow I=\int{\dfrac{\sin x}{\sin x-\cos x}dx}$
Multiplying the numerator and denominator by $\sin x+\cos x$, we get
$\Rightarrow \int{\dfrac{\sin x\left( \sin x+\cos x \right)}{\sin x-\cos x\left( \sin x+\cos x \right)}dx}$
Simplifying the above integrating function, we get
$\Rightarrow \int{\dfrac{{{\sin }^{2}}x+\sin x\cos x}{\left( \sin x-\cos x \right)\left( \sin x+\cos x \right)}dx}$
Putting the identity of ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$ in the denominator, we get
$\Rightarrow \int{\dfrac{{{\sin }^{2}}x+\sin x\cos x}{{{\sin }^{2}}x-{{\cos }^{2}}x}dx}$
Split it into two parts, we get
$\Rightarrow \int{\dfrac{{{\sin }^{2}}x}{{{\sin }^{2}}x-{{\cos }^{2}}x}dx+\int{\dfrac{\sin x\cos x}{{{\sin }^{2}}x-{{\cos }^{2}}x}dx}}$
Using the identity of trigonometry i.e.
${{\cos }^{2}}x-{{\sin }^{2}}x=\cos 2x$
$\sin 2x=2\sin x\cos x$
${{\sin }^{2}}x=\dfrac{1-\cos 2x}{2}$
Applying these identity in the solved integral, we get
$\Rightarrow -\int{\dfrac{1-\cos 2x}{2\cos 2x}dx-\int{\dfrac{\sin 2x}{2\cos 2x}dx}}$
$\Rightarrow -\int{\dfrac{\sec 2x}{2}dx+\int{\dfrac{1}{2}dx-\int{\dfrac{\tan 2x}{2}dx}}}$
Using,
$\int{\tan \left( x \right)dx}=\log \left| \sec x \right|+C$
$\int{\sec \left( x \right)dx=\log \left| \sec x+\tan x \right|}+C$
$\int{dx=x+C}$
After applying these identities of trigonometry, we get
$\Rightarrow \dfrac{\log \left| \sec 2x+\tan 2x \right|}{4}+\dfrac{x}{2}-\dfrac{\log \left| \sec 2x \right|}{4}+C$
Therefore,
$\Rightarrow \int{\dfrac{\sin x}{\sin x-\cos x}dx}=\dfrac{\log \left| \sec 2x+\tan 2x \right|}{4}+\dfrac{x}{2}-\dfrac{\log \left| \sec 2x \right|}{4}+C$

Note: In order to solve the given question, we used trigonometric identities to simplify the given function and then split it into two terms. We split the function inside the integral only because the operation between both the functions is addition. If there is a mathematical operation of multiplication, then here we integrate the given function by using the integration by parts. We should remember the trigonometric identities, this would make it easier to solve the question. There are also many other methods for integration. These are integration by substitution and integration by partial fractions. You should always remember all the methods for integration so that we can easily choose which method is suitable for solving the particular type of question. We should do all the calculations carefully and explicitly to avoid making errors.