Question

Find the value of $\int\limits_0^{\dfrac{\pi }{2}} {\left( {\dfrac{{{{\cos }^5}x}}{{{{\sin }^5}x + {{\cos }^5}x}}} \right)} dx$?

Answer 1

To solve and take out the integral of this question , we need to solve it step by step. Here we are going to let the whole question be equated to “I“, with the help of the concept of integration we will perform some calculations and apply the formulae of integration to simplify the given question . Also with the help of the concept of limits integration $\int\limits_b^a {f(x)}$ The function $f(x)$ is called the integrand, and the variable $x$ is the variable of integration. The numbers $a$ and $b$ are called the limits of integration with a referred to as the lower limit of integration while b is referred to as the upper limit of integration. and its formulae which we are going to apply in this question , will make it easier to integrate and get the desired value. Some basic trigonometry formulae will come into existence while solving the integration part.

Complete step by step solution:
We are given a expression $\int\limits_0^{\dfrac{\pi }{2}} {\left( {\dfrac{{{{\cos }^5}x}}{{{{\sin }^5}x + {{\cos }^5}x}}} \right)} dx$ and we have to calculate its value .
The trigonometric formula used here is –
$\Rightarrow \cos \left( {\dfrac{\pi }{2} - \theta } \right) = \sin \theta \\\ \Rightarrow \sin \left( {\dfrac{\pi }{2} - \theta } \right) = \cos \theta \\\$
First , we will Let the expression equal to “ I “ .
$I = \int\limits_0^{\dfrac{\pi }{2}} {\left( {\dfrac{{{{\cos }^5}x}}{{{{\sin }^5}x + {{\cos }^5}x}}} \right)} dx$ ---------- equation (1)
Now apply The integration formula here –
$\int\limits_0^a {f(x)} dx = \int\limits_0^a {f(a - x)} dx$
We get-
$I = \int\limits_0^{\dfrac{\pi }{2}} {\left( {\dfrac{{{{\left[ {\cos \left( {\dfrac{\pi }{2} - x} \right)} \right]}^5}}}{{{{\left[ {\sin \left( {\dfrac{\pi }{2} - x} \right)} \right]}^5} + {{\left[ {\cos \left( {\dfrac{\pi }{2} - x} \right)} \right]}^5}}}} \right)} dx$
Now here we are going to use the trigonometric formula and substitute here –
$\Rightarrow \cos \left( {\dfrac{\pi }{2} - \theta } \right) = \sin \theta \\\ \Rightarrow \sin \left( {\dfrac{\pi }{2} - \theta } \right) = \cos \theta \\\$

On applying , We get –
$I = \int\limits_0^{\dfrac{\pi }{2}} {\left( {\dfrac{{{{\sin }^5}x}}{{{{\cos }^5}x + {{\sin }^5}x}}} \right)} dx$ ---------- equation (2)
Now we are going to add the equation (1) and equation (2)

$$\Rightarrow I = \int\limits_0^{\dfrac{\pi }{2}} {\left( {\dfrac{{{{\cos }^5}x}}{{{{\sin }^5}x + {{\cos }^5}x}}} \right)} dx + I = \int\limits_0^{\dfrac{\pi }{2}} {\left( {\dfrac{{{{\sin }^5}x}}{{{{\cos }^5}x + {{\sin }^5}x}}} \right)} dx \\\ \Rightarrow 2I = \int\limits_0^{\dfrac{\pi }{2}} {\left( {\dfrac{{{{\cos }^5}x + {{\sin }^5}x}}{{{{\sin }^5}x + {{\cos }^5}x}}} \right)} dx \\\ \Rightarrow 2I = \int\limits_0^{\dfrac{\pi }{2}} {\left( 1 \right)} dx \\\$$

Now , we know the limits integration formula so , we will apply the same here –
$\Rightarrow \int\limits_b^a {f(x)} dx = f(a) - f(b)$
Applying the same , we get –
\Rightarrow 2I = \int\limits_0^{\dfrac{\pi }{2}} {\left( 1 \right)} dx \\\
$\Rightarrow 2I= x$

$$\Rightarrow I = \dfrac{\pi }{4} \\\$$

Therefore, the value of the $\int\limits_0^{\dfrac{\pi }{2}} {\left( {\dfrac{{{{\cos }^5}x}}{{{{\sin }^5}x + {{\cos }^5}x}}} \right)} dx$ is $\dfrac{\pi }{4}$.

Note:
1. Use the standard formula carefully while evaluating the integrals.
2. Indefinite integral: Let $f(x)$ be a function. Then the family of all its primitives (or antiderivatives) is called the indefinite integral of $f(x)$ and is denoted by $\int {f(x)} dx$
3. The symbol $\int\limits_b^a {f(x)}$ is read as the definite integral of $f(x)$ with respect to x.
4. C is known as the constant of integration.
5. In our question, the given function is a definite integral, because it is evaluated over a certain interval. Although the constant is strictly not necessary, because it will be subtracted when the integral is evaluated, it is good practice to keep the constant of integration.