Question

Find the value of $\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{\cos x-\sin x}{1+\sin x\cos x}dx}$.
A. $2$
B. $-2$
C. $0$
D. None of these

Answer 1

Hint: In this question, we have $\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{\cos x-\sin x}{1+\sin x\cos x}dx}$. So, we will consider this equation as 1. Now we will use the identity $\int\limits_{0}^{a}{f(x)dx=\int\limits_{0}^{a}{f(a-x)dx}}$. After that, we will get equation 2. Now we will add equations 1 and 2 and then we will use the Cofunction identities. At last, we will obtain the final result.



Formula Used:For the given problem, we will use the following formulas:
1) $\int_{0}^{a}{f\left( x \right)dx=}\int_{0}^{a}{f\left( a-x \right)dx}$
2) $\sin \left( \dfrac{\pi }{2}-x \right)=\cos x,\cos \left( \dfrac{\pi }{2}-x \right)=\sin x$



Complete step by step solution:We are given an integral as $\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{\cos x-\sin x}{1+\sin x\cos x}dx}$.
Let us consider $I=\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{\cos x-\sin x}{1+\sin x\cos x}dx}$ ….... (1)
First, we will use the identity $\int_{0}^{a}{f\left( x \right)dx=}\int_{0}^{a}{f\left( a-x \right)dx}$.
$I=\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{\cos \left( \dfrac{\pi }{2}-x \right)-\sin \left( \dfrac{\pi }{2}-x \right)}{1+\sin \left( \dfrac{\pi }{2}-x \right)\cos \left( \dfrac{\pi }{2}-x \right)}dx}$
As we know that the cofunction identities of sine and cosine:
$\sin \left( \dfrac{\pi }{2}-x \right)=\cos x,\cos \left( \dfrac{\pi }{2}-x \right)=\sin x$
So, the integral becomes
$I=\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{\sin x-\cos x}{1+\cos x\sin x}dx}$ ….... (2)
Now, we are adding equations (1) and (2).
$\begin{align}
  & \Rightarrow I+I=\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{\cos x-\sin x}{1+\sin x\cos x}dx+\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{\sin x-\cos x}{1+\cos x\sin x}}}dx \\
 & \Rightarrow 2I=\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{\cos -\sin x+\sin x-\cos x}{1+\sin x\cos x}}dx \\
\end{align}$
By cancelling the same terms, we get
$2I=\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{0}{1+\sin x\cos x}}dx$
We will further integrate from $0$ to $\dfrac{\pi }{2}$.
$\begin{align}
  & \Rightarrow 2I=0 \\
 & \Rightarrow I=0 \\
\end{align}$
As a result, integral of $\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{\cos x-\sin x}{1+\sin x\cos x}dx}$ is $0$.



Option ‘C’ is correct


Note: As the functions change, it is not essential to integrate taking limits directly $0$ to $\dfrac{\pi }{2}$. We must avoid errors and confusion while dealing with sign changes that occurred during integration. We need to adopt the identities that make the problems easier. Students are advised to write the correct identity and then solve the problem. Don’t choose the identity that makes the question more complicated.