Question

Evaluate the Integral $\int\limits_{\dfrac{-\pi }{4}}^{\dfrac{\pi }{4}}{\dfrac{1}{1-\sin x}}dx$.

Answer 1

This is the question of integration. First, we will simplify the integration and then we will integrate them and after that, we will put the limits in the integration. At first, we will put the upper limit and after that, we will put the lower limit, and then we will subtract them to obtain the result.

Complete step-by-step solution:
In mathematics, integration is of two types, indefinite integral and definite integral. In an indefinite integral, limits are not given but in a definite integral, limits are given and we have to solve the question by first putting the upper limit and then the lower limit.
Before solving integration you should know about the formulas of integration and if it is trigonometric integration then trigonometric formulas should also be known.
In the above question, we have to integrate $\int\limits_{\dfrac{-\pi }{4}}^{\dfrac{\pi }{4}}{\dfrac{1}{1-\sin x}}dx$.
This is the question of integration and limits are also given in it. First, we will simplify the integration and after applying the limits, we will get the results.
Let, $I=\int\limits_{\dfrac{-\pi }{4}}^{\dfrac{\pi }{4}}{\dfrac{1}{1-\sin x}dx}$…....eq(1)
In eq(1), we will multiply the numerator and denominator by $1+\sin x$, which is as shown below.
$I=\int\limits_{\dfrac{-\pi }{4}}^{\dfrac{\pi }{4}}{\dfrac{1+\sin x}{(1-\sin x)(1+\sin x)}}dx$…..eq(2)
we know that, $(a-b)(a+b)={{a}^{2}}-{{b}^{2}}$
So, $\begin{aligned} & (1-\sin x)(1+\sin x)=1-{{\sin }^{2}}x \\\ & \\\ \end{aligned}$
On putting this in eq(2), we get the following results.
$I=\int\limits_{\dfrac{-\pi }{4}}^{\dfrac{\pi }{4}}{\dfrac{1+\sin x}{1-{{\sin }^{2}}x}}dx$……eq(3)
According to the identity of trigonometry,
${{\sin }^{2}}x+{{\cos }^{2}}x=1$,
$\Rightarrow {{\cos }^{2}}x=1-{{\sin }^{2}}x$
On putting the value of ${{\cos }^{2}}x$ in eq(3), we get the following results as shown below.
$I=\int\limits_{\dfrac{-\pi }{4}}^{\dfrac{\pi }{4}}{\dfrac{1+\sin x}{{{\cos }^{2}}x}}dx$…….eq(4)
On taking the denominator separately below the numerator.
$I=\int\limits_{\dfrac{-\pi }{4}}^{\dfrac{\pi }{4}}{\left( \dfrac{1}{{{\cos }^{2}}x}+\dfrac{\sin x}{{{\cos }^{2}}x} \right)}dx$,
$I=\int\limits_{\dfrac{-\pi }{4}}^{\dfrac{\pi }{4}}{\left( \dfrac{1}{{{\cos }^{2}}x}+\dfrac{\sin x}{\cos x\cos x} \right)}dx$……eq(5)
According to the formula of trigonometry,
$\dfrac{1}{{{\cos }^{2}}x}={{\sec }^{2}}x$
$\dfrac{1}{\cos x}=\sec x$
$\dfrac{\sin x}{\cos x}=\tan x$
On putting all these values in eq(5), we get the following results.
$I=\int\limits_{\dfrac{-\pi }{4}}^{\dfrac{\pi }{4}}{\left( {{\sec }^{2}}x+\sec x\tan x \right)}dx$
On integrating the above expression separately.
$I=\int\limits_{\dfrac{-\pi }{4}}^{\dfrac{\pi }{4}}{{{\sec }^{2}}x}dx+\int\limits_{\dfrac{-\pi }{4}}^{\dfrac{\pi }{4}}{\sec x\tan xdx}$…….eq(6)
We know that,
$\int{{{\sec }^{2}}xdx}=\tan x$
$\int{\sec x\tan xdx}=\sec x$
On putting these values in eq(5),
$I=\left[ \tan x \right]_{\dfrac{-\pi }{4}}^{\dfrac{\pi }{4}}+\left[ \sec x\tan x \right]_{\dfrac{-\pi }{4}}^{\dfrac{\pi }{4}}$
After integrating we will apply the limits on the integration
$I=\left[ \tan \dfrac{\pi }{4}-\tan \left( \dfrac{-\pi }{4} \right) \right]+\left[ \sec \dfrac{\pi }{4}-\sec \left( \dfrac{-\pi }{4} \right) \right]$….eq(6)
We know that,
$\tan \dfrac{\pi }{4}=1$
And we also know that,
$\tan (-\theta )=-\tan \theta$
Therefore, $\tan \left( \dfrac{-\pi }{4} \right)=-\tan \dfrac{\pi }{4}$
We also know that,
$\sec \dfrac{\pi }{4}=\sqrt{2}$
And, $\sec (-\theta )=\sec \theta$
Therefore, $\sec \left( \dfrac{-\pi }{4} \right)=\sec \dfrac{\pi }{4}$
On putting all these values in eq(6)
$I=\left[ \tan \dfrac{\pi }{4}+\tan \dfrac{\pi }{4} \right]+\left[ \sec \dfrac{\pi }{4}-\sec \dfrac{\pi }{4} \right]$
Now on putting the respective values in the above equation, we get.

& I=\left[ 1+1 \right]+\left[ \sqrt{2}-\sqrt{2} \right] \\\ & \Rightarrow I=2 \\\ \end{aligned}$$ **Hence the result of the integration is $$2$$. $$\int\limits_{\dfrac{-\pi }{4}}^{\dfrac{\pi }{4}}{\dfrac{1}{1-\sin x}dx=2}$$** **Note:** Integration and differentiation are both the topics of calculus in mathematics. Differentiation is the reverse of integration. With the help of integration, we can also find out the values of the area, volume, displacement, etc. The integration of the sum of two functions is obtained by adding the integration of these functions individually.