Question

Find $\int {\dfrac{{\sin 2x + 1}}{{1 - \sin 2x}}} dx$.

Answer 1

Given trigonometric function is $\dfrac{{\sin 2x + 1}}{{1 - \sin 2x}}$. We have to integrate the function with respect to x. In order to approach our solution, first, we can simplify the given function to smaller form so that it becomes easy to integrate. We can use a few trigonometric identities to simplify the function. Such as
$\Rightarrow \sin 2x = \dfrac{{2\tan x}}{{1 + {{\tan }^2}x}}$
$\Rightarrow 1 + {\tan ^2}x = {\sec ^2}x$
After the simplification, we can easily integrate the obtained function with respect to x by using the substitution method.

Complete step by step answer:
The trigonometric function is $\dfrac{{\sin 2x + 1}}{{1 - \sin 2x}}$ . We have to integrate the function with respect to x. Let
$\Rightarrow I = \int {\dfrac{{\sin 2x + 1}}{{1 - \sin 2x}}} dx$
We can write the above expression as,
$\Rightarrow I = \int {\dfrac{{1 + \sin 2x}}{{1 - \sin 2x}}} dx$
Taking a minus sign twice in the above expression, gives
$\Rightarrow I = - \int {\dfrac{{ - 1 - \sin 2x}}{{1 - \sin 2x}}} dx$

Writing $- 1 = {\text{ }}1 - 2$ , we get
$\Rightarrow I = - \int {\dfrac{{1 - \sin 2x - 2}}{{1 - \sin 2x}}} dx$
We can write it as,
$\Rightarrow I = - \int {1 - \dfrac{2}{{1 - \sin 2x}}} dx$
That gives,
$\Rightarrow I = - \int {dx + 2\int {\dfrac{{dx}}{{1 - \sin 2x}}} }$
Therefore,
$\Rightarrow I = - x + 2\int {\dfrac{{dx}}{{1 - \sin 2x}}}$ ...(1)
Now, let
$\Rightarrow {I_2} = 2\int {\dfrac{{dx}}{{1 - \sin 2x}}}$
Using the trigonometric identities,
$\Rightarrow \sin 2x = \dfrac{{2\tan x}}{{1 + {{\tan }^2}x}}$

And
$\Rightarrow 1 + {\tan ^2}x = {\sec ^2}x$
We can write,
$\Rightarrow \sin 2x = \dfrac{{2\tan x}}{{{{\sec }^2}x}}$
Now putting $\sin 2x = \dfrac{{2\tan x}}{{{{\sec }^2}x}}$ in ${I_2} = 2\int {\dfrac{{dx}}{{1 - \sin 2x}}}$ , we get
$\Rightarrow {I_2} = 2\int {\dfrac{{dx}}{{1 - \dfrac{{2\tan x}}{{{{\sec }^2}x}}}}}$
We can write the above expression as,
$\Rightarrow {I_2} = 2\int {\dfrac{{dx}}{{\dfrac{{{{\sec }^2}x}}{{{{\sec }^2}x}} - \dfrac{{2\tan x}}{{{{\sec }^2}x}}}}}$
So we get,
$\Rightarrow {I_2} = 2\int {\dfrac{{dx}}{{\dfrac{{{{\sec }^2}x - 2\tan x}}{{{{\sec }^2}x}}}}}$
$\Rightarrow {I_2} = 2\int {\dfrac{{{{\sec }^2}x}}{{{{\sec }^2}x - 2\tan x}}dx}$

Again using the trigonometric identity $1 + {\tan ^2}x = {\sec ^2}x$ , we can write the above expression as,
$\Rightarrow {I_2} = 2\int {\dfrac{{{{\sec }^2}x}}{{1 + {{\tan }^2}x - 2\tan x}}dx}$
Now using the formula ${a^2} + {b^2} + 2ab = {\left( {a + b} \right)^{}}$ in above expression, we get
$\Rightarrow {I_2} = 2\int {\dfrac{{{{\sec }^2}x}}{{{{\left( {\tan x - 1} \right)}^2}}}dx}$
Now, substitute $\tan x - 1 = u$
Since,
$\Rightarrow u = \tan x - 1$
Therefore, differentiating both sides, gives
$\Rightarrow du = {\sec ^2}xdx + 0$
Hence,
$\Rightarrow \dfrac{{du}}{{dx}} = {\sec ^2}x$

Now putting $u = \tan x - 1$ and $du = {\sec ^2}xdx$ in ${I_2} = 2\int {\dfrac{{{{\sec }^2}x}}{{{{\left( {\tan x - 1} \right)}^2}}}dx}$
We get,
$\Rightarrow {I_2} = 2\int {\dfrac{{du}}{{{u^2}}}}$
Therefore,
$\Rightarrow {I_2} = - \dfrac{2}{u} + c$
Since $u = \tan x - 1$ , we get
$\Rightarrow {I_2} = - \dfrac{2}{{\tan x - 1}} + c$
Where c is the integration constant.
Now putting the value ${I_2} = - \dfrac{2}{{\tan x - 1}} + c$ in ...(1) , we get
$\Rightarrow I = - x - \dfrac{2}{{\tan x - 1}} + c$
Therefore, we get
$\therefore \int {\dfrac{{\sin 2x + 1}}{{1 - \sin 2x}}} dx = - x - \dfrac{2}{{\tan x - 1}} + c$
That is the required integration of the given expression.

Therefore, the value of the function $\int {\dfrac{{\sin 2x + 1}}{{1 - \sin 2x}}} dx$ is $- x - \dfrac{2}{{\tan x - 1}} + c$ , where c is the integration constant.

Note: There are various more methods of integration for different functions and they are used wherever they are suitable and make the integration easier to evaluate. Some of the various common methods of integration are, namely:
-Integration by substitution.
-Integration by parts.
-Integration using trigonometric identities.
-Integration of some particular functions.
-Integration by partial fraction.
These methods can be used to integrate different functions.