Question

Integrate the following \[\int{\dfrac{dx}{\cos x-\sin x}}\]

Answer 1

Hint: From the question, it is clear that we should find the value of \[\int{\dfrac{dx}{\cos x-\sin x}}\]. Let us assume the value of \[\int{\dfrac{dx}{\cos x-\sin x}}\] is equal to I. Now let us multiply and divide with \[\dfrac{1}{\sqrt{2}}\] on R.H.S. We know that \[\cos \dfrac{\pi }{4}=\sin \dfrac{\pi }{4}=\dfrac{1}{\sqrt{2}}\]. By using \[\cos \dfrac{\pi }{4}=\sin \dfrac{\pi }{4}=\dfrac{1}{\sqrt{2}}\], we should write the denominator in the form of \[\cos A\cos B-\sin A\sin B\]. We know that \[\cos \left( A+B \right)=\cos A\cos B-\sin A\sin B\]. Now by using this formula, we should write the denominator in the form of \[\cos \theta \]. We know that \[\sec \theta =\dfrac{1}{\cos \theta }\]. We know that \[\int{\sec \theta }=\ln \left| \tan \theta +\sec \theta \right|\]. By using this formula, we can find the value of \[\int{\dfrac{dx}{\cos x-\sin x}}\].

Complete step-by-step answer:
From the question, it is clear that we should find the value of \[\int{\dfrac{dx}{\cos x-\sin x}}\].
Let us assume the value of \[\int{\dfrac{dx}{\cos x-\sin x}}\] is equal to I.
\[I=\int{\dfrac{dx}{\cos x-\sin x}}\]
Now let us multiply and divide with \[\dfrac{1}{\sqrt{2}}\].
\[\begin{align}
  & \Rightarrow I=\dfrac{1}{\sqrt{2}}\int{\dfrac{dx}{\dfrac{1}{\sqrt{2}}\left( \cos x-\sin x \right)}} \\
 & \Rightarrow I=\dfrac{1}{\sqrt{2}}\int{\dfrac{dx}{\dfrac{1}{\sqrt{2}}\cos x-\dfrac{1}{\sqrt{2}}\sin x}} \\
\end{align}\]
We know that \[\cos \dfrac{\pi }{4}=\sin \dfrac{\pi }{4}=\dfrac{1}{\sqrt{2}}\].
\[\Rightarrow I=\dfrac{1}{\sqrt{2}}\int{\dfrac{dx}{cos\dfrac{\pi }{4}\cos x-\sin \dfrac{\pi }{4}\sin x}}\]
We know that \[\cos \left( A+B \right)=\cos A\cos B-\sin A\sin B\].
\[\Rightarrow I=\dfrac{1}{\sqrt{2}}\int{\dfrac{dx}{cos\left( x+\dfrac{\pi }{4} \right)}}\]
We know that \[\sec \theta =\dfrac{1}{\cos \theta }\].
\[\Rightarrow I=\dfrac{1}{\sqrt{2}}\int{\sec \left( x+\dfrac{\pi }{4} \right)dx}.......(1)\]
Let us assume \[y=x+\dfrac{\pi }{4}.....(2)\].
Now let us differentiate equation (2) with respect to x on both sides.
\[\begin{align}
  & \Rightarrow \dfrac{dy}{dx}=\dfrac{d}{dx}\left( x+\dfrac{\pi }{4} \right) \\
 & \Rightarrow \dfrac{dy}{dx}=\dfrac{dx}{dx}+\dfrac{d}{dx}\left( \dfrac{\pi }{4} \right) \\
 & \Rightarrow \dfrac{dy}{dx}=1 \\
\end{align}\]
Now we will apply cross multiplication.
\[\Rightarrow dy=dx......(3)\]
Now let us substitute equation (2) and equation (3) in equation (1), then we get
\[\Rightarrow I=\dfrac{1}{\sqrt{2}}\int{\operatorname{secy}dy}.......(4)\]
We know that \[\int{\sec \theta }=\ln \left| \tan \theta +\sec \theta \right|\].
\[\Rightarrow I=\dfrac{1}{\sqrt{2}}\left( \ln \left| \operatorname{tany}+\operatorname{secy} \right| \right)+C..........(5)\]
Now let us substitute equation (2) in equation (5), then we get
\[\Rightarrow I=\dfrac{1}{\sqrt{2}}\left( \ln \left| \tan \left( x+\dfrac{\pi }{4} \right)+\sec \left( x+\dfrac{\pi }{4} \right) \right| \right)+C.......(6)\]
From equation (6), we can say that
\[\Rightarrow \int{\dfrac{dx}{\cos x-\sin x}}=\dfrac{1}{\sqrt{2}}\left( \ln \left| \tan \left( x+\dfrac{\pi }{4} \right)+\sec \left( x+\dfrac{\pi }{4} \right) \right| \right)+C\]

Note: Students may assume a misconception that \[\int{\sec \theta }=\ln \left| \tan \theta -\sec \theta \right|\]. If this formula is applied, then we get the value of I is equal to \[\dfrac{1}{\sqrt{2}}\left( \ln \left| \tan \left( x+\dfrac{\pi }{4} \right)-\sec \left( x+\dfrac{\pi }{4} \right) \right| \right)+C\]. Then, we get the value of \[\int{\dfrac{dx}{\cos x-\sin x}}\] is equal to \[\dfrac{1}{\sqrt{2}}\left( \ln \left| \tan \left( x+\dfrac{\pi }{4} \right)-\sec \left( x+\dfrac{\pi }{4} \right) \right| \right)+C\]. But we know that the value of \[\int{\dfrac{dx}{\cos x-\sin x}}\] is equal to \[\dfrac{1}{\sqrt{2}}\left( \ln \left| \tan \left( x+\dfrac{\pi }{4} \right)+\sec \left( x+\dfrac{\pi }{4} \right) \right| \right)+C\]. So, this misconception should be avoided to get an accurate result.