Question

Find the value of \[\int {\dfrac{{\cos x - \sin x}}{{\sqrt {8 - \sin 2x} }}} dx\].

Answer 1

Hint: We will assume $\cos x + \sin x = t$ and find dx in terms of t and dt as well. We will then put in these in the given integral and will get an easy integral in t and dt. On solving that, we will eventually put in the value of t as we assumed earlier.

Complete step-by-step answer:
We will first of all assume that $\cos x + \sin x = t$. …………….(1)
Now, differentiating both sides of (1), we will get:-
\[ \Rightarrow \dfrac{d}{{dx}}\left( {\cos x + \sin x} \right) = \dfrac{{dt}}{{dx}}\]
We can rewrite it as follows:
\[ \Rightarrow \dfrac{d}{{dx}}\left( {\cos x} \right) + \dfrac{d}{{dx}}\left( {\sin x} \right) = \dfrac{{dt}}{{dx}}\] ………..(2)
We know that $\dfrac{d}{{dx}}\cos x = - \sin x$ and $\dfrac{d}{{dx}}\sin x = \cos x$
Using these in (2), we will get:-
\[ \Rightarrow - \sin x + \cos x = \dfrac{{dt}}{{dx}}\]
Taking dx from the denominator in RHS to numerator in LHS, we will get:-
\[ \Rightarrow \left( { - \sin x + \cos x} \right)dx = dt\]
We can rewrite it as follows:-
\[ \Rightarrow \left( {\cos x - \sin x} \right)dx = dt\] ……………..(3)
Now, squaring (1), we will get as follows:-
$ \Rightarrow {\left( {\cos x + \sin x} \right)^2} = {t^2}$
Now, we will use the formula ${(a + b)^2} = {a^2} + {b^2} + 2ab$. We will then get as follows:-
\[ \Rightarrow {\cos ^2}x + {\sin ^2}x + 2\sin x\cos x = {t^2}\]
Now, we also know that ${\sin ^2}\theta + {\cos ^2}\theta = 1$ for all values of $\theta \in \mathbb{R}$.
Hence, we get:
\[ \Rightarrow 1 + 2\sin x\cos x = {t^2}\]
We know that \[2\sin x\cos x = \sin 2x\].
\[ \Rightarrow 1 + \sin 2x = {t^2}\] ……….(4)
Putting (3) and (4) in the given expression, we will thus get:-
\[ \Rightarrow \int {\dfrac{{\cos x - \sin x}}{{\sqrt {8 - \sin 2x} }}} dx = \int {\dfrac{{\cos x - \sin x}}{{\sqrt {9 - (1 + \sin 2x)} }}} dx = \int {\dfrac{{dt}}{{\sqrt {9 - {t^2}} }}} \]
We know that \[\int {\dfrac{{dx}}{{\sqrt {{a^2} - {x^2}} }} = {{\sin }^{ - 1}}\dfrac{x}{a} + C} \]
Therefore, we get \[ \Rightarrow \int {\dfrac{{\cos x - \sin x}}{{\sqrt {8 - \sin 2x} }}} dx = \int {\dfrac{{dt}}{{\sqrt {9 - {t^2}} }}} = {\sin ^{ - 1}}\dfrac{t}{3} + C\]
Putting (1) in this, we will get:-
\[ \Rightarrow \int {\dfrac{{\cos x - \sin x}}{{\sqrt {8 - \sin 2x} }}} dx = {\sin ^{ - 1}}\dfrac{t}{3} + C = {\sin ^{ - 1}}\left( {\dfrac{{\cos x + \sin x}}{3}} \right) + C\].

Note: The students may also try approaching the problem directly which means without any modifications but that may turn out to be very chaotic and confusing. SO, modifying our problems into simpler ones help us to easily integrate them.
Integration is the reverse of derivatives. Derivative refers to the rate of change. Integral is basically equivalent to summation. We use summation when working with discrete values, but when we work with continuous values, we use integral. In integral, you basically are cutting an area into very small rectangles and summing up all the areas of all the rectangles where f(x) is the curve which is equal to length of the rectangle and dx is the width of the triangle and multiplying them gives us the required area.