Question

Find the value of given integration \[\int_{ - \pi }^\pi {\dfrac{{{{\sin }^4}x}}{{{{\sin }^4}x + {{\cos }^4}x}}dx} \].
A. \[\dfrac{\pi }{4}\]
B. \[\dfrac{\pi }{2}\]
C. \[\dfrac{{3\pi }}{2}\]
D. \[2\pi \]
E. \[\pi \]

Answer 1

Hint: To find the value of the integration first convert the function in integrable form. Then integrate the function and apply the limits.

Formula used: If the function is even then,
\[\int_{ - a}^a {f\left( x \right)dx} = 2\int_0^a {f\left( x \right)} \]
And
\[\int_0^{2a} {f\left( x \right)dx} = 2\int_0^a {f\left( x \right)} \] if \[f\left( {2a - x} \right) = f\left( x \right)\]
Also,
\[\int_a^b {f\left( x \right)dx} = \int_a^b {f\left( {a + b - x} \right)dx} \]
\[\begin{array}{l}\sin \left( {\dfrac{\pi }{2} - \theta } \right) = \cos \theta \\\cos \left( {\dfrac{\pi }{2} - \theta } \right) = \sin \theta \end{array}\]

Complete step by step solution: The given integral is \[\int_{ - \pi }^\pi {\dfrac{{{{\sin }^4}x}}{{{{\sin }^4}x + {{\cos }^4}x}}dx} \].
Consider \[I = \int_{ - \pi }^\pi {\dfrac{{{{\sin }^4}x}}{{{{\sin }^4}x + {{\cos }^4}x}}dx} \].
The function \[\dfrac{{{{\sin }^4}x}}{{{{\sin }^4}x + {{\cos }^4}x}}\]is a even function so integral becomes as follows.
\[I = 2\int_0^\pi {\dfrac{{{{\sin }^4}x}}{{{{\sin }^4}x + {{\cos }^4}x}}dx} \]
Further we can simplify the integral as follows.
\[I = 2 \times 2\int_0^{\dfrac{\pi }{2}} {\dfrac{{{{\sin }^4}x}}{{{{\sin }^4}x + {{\cos }^4}x}}dx} \]
So, the integral becomes as follows.
\[I = 4\int_0^{\dfrac{\pi }{2}} {\dfrac{{{{\sin }^4}x}}{{{{\sin }^4}x + {{\cos }^4}x}}dx} \] …(1)
Apply the property to simplify the function.
\[I = 4\int_0^{\dfrac{\pi }{2}} {\dfrac{{{{\sin }^4}\left( {\dfrac{\pi }{2} + 0 - x} \right)}}{{{{\sin }^4}\left( {\dfrac{\pi }{2} + 0 - x} \right) + {{\cos }^4}\left( {\dfrac{\pi }{2} + 0 - x} \right)}}dx} \]
Hence the integral becomes like this.
\[I = 4\int_0^{\dfrac{\pi }{2}} {\dfrac{{{{\sin }^4}\left( {\dfrac{\pi }{2} - x} \right)}}{{{{\sin }^4}\left( {\dfrac{\pi }{2} - x} \right) + {{\cos }^4}\left( {\dfrac{\pi }{2} - x} \right)}}dx} \]
Here, apply the formula to further simplify.
\[I = 4\int_0^{\dfrac{\pi }{2}} {\dfrac{{{{\cos }^4}x}}{{{{\cos }^4}x + {{\sin }^4}x}}dx} \] …(2)
Now add equation 1 and equation 2.
\[I + I = 4\int_0^{\dfrac{\pi }{2}} {\dfrac{{{{\sin }^4}x}}{{{{\sin }^4}x + {{\cos }^4}x}}dx} + 4\int_0^{\dfrac{\pi }{2}} {\dfrac{{{{\cos }^4}x}}{{{{\cos }^4}x + {{\sin }^4}x}}dx} \]
Simplify the equation.
\[\begin{array}{l}2I = 4\int_0^{\dfrac{\pi }{2}} {\dfrac{{{{\sin }^4}x + {{\cos }^4}x}}{{{{\sin }^4}x + {{\cos }^4}x}}dx} \\2I = 4\int_0^{\dfrac{\pi }{2}} {1dx} \end{array}\]
Evaluate the integral.
\[2I = 4\left[ x \right]_0^{\dfrac{\pi }{2}}\]
Substitute the limits and simplify it
\[\begin{array}{l}2I = 4\left[ {\dfrac{\pi }{2} - 0} \right]\\2I = 2\pi \end{array}\]
Divide by \[2\] on both sides of the equation.
\[I = \pi \]
Hence the value of \[\int_{ - \pi }^\pi {\dfrac{{{{\sin }^4}x}}{{{{\sin }^4}x + {{\cos }^4}x}}dx} \] is \[\pi \].

Thus, Option (E) is correct.

Note: The common mistake happens while solving this type of question is students solve the function which leads the question into complication. Instead of that as the function, apply the formula and solve the integral.