Question

Integrate $\int {\dfrac{{{{\cos }^2}x - {{\sin }^2}x}}{{\sqrt {1 + \cos 4x} }}dx} $.

Answer 1

Hint: First, use the identity, $\cos 2x = {\cos ^2}x - {\sin ^2}x$ in the numerator. After that use the identity, $\cos 2x = 2{\cos ^2}x - 1$ in the denominator. After that cancel out square with square root in the denominator. Then, cancel out the common terms from the numerator and denominator. Then integrate the term to get the desired result.

Complete step-by-step solution:
It is given in the question that we have to integrate $\int {\dfrac{{{{\cos }^2}x - {{\sin }^2}x}}{{\sqrt {1 + \cos 4x} }}dx} $.
To start integrating the given function, we have to name the integral. Therefore, we consider it as
$ \Rightarrow I = \int {\dfrac{{{{\cos }^2}x - {{\sin }^2}x}}{{\sqrt {1 + \cos 4x} }}dx} $
Now use the identity, $\cos 2x = {\cos ^2}x - {\sin ^2}x$ in the numerator.
$ \Rightarrow I = \int {\dfrac{{\cos 2x}}{{\sqrt {1 + \cos 4x} }}dx} $
Now use the identity, $\cos 2x = 2{\cos ^2}x - 1$ in the denominator.
$ \Rightarrow I = \int {\dfrac{{\cos 2x}}{{\sqrt {2{{\cos }^2}2x} }}dx} $
Write the denominator part in the square,
$ \Rightarrow I = \int {\dfrac{{\cos 2x}}{{\sqrt {{{\left( {\sqrt 2 \cos 2x} \right)}^2}} }}dx} $
Now cancel out the square with the square root,
$ \Rightarrow I = \int {\dfrac{{\cos 2x}}{{\sqrt 2 \cos 2x}}dx} $
Now cancel out the common factor from the numerator and denominator,
$ \Rightarrow I = \int {\dfrac{1}{{\sqrt 2 }}dx} $
Take out constant term outside the integration,
$ \Rightarrow I = \dfrac{1}{{\sqrt 2 }}\int {1dx} $
Now integrate the term,
$ \Rightarrow I = \dfrac{x}{{\sqrt 2 }}$

Hence, the integration of $\int {\dfrac{{{{\cos }^2}x - {{\sin }^2}x}}{{\sqrt {1 + \cos 4x} }}dx} $ is $\dfrac{x}{{\sqrt 2 }}$.

Note: Integration is often wont to find areas, volumes, central points, and lots of useful things. it's often wont to find the world under the graph of a function. The symbol for "Integral" may be a stylish "S" (for "Sum", the thought of summing slices).
This question is often solved in only a couple of steps by using basic integration rules and formulas. So, the students must confirm to memorize all the essential integration formulae. This may save time for solving the question. Also, integration is the reverse of differentiation. So, the students can even check if the obtained answer is correct by differentiating the obtained result with reference to x. If the student gets the question on the solution after differentiation, the answer would even be correct.