Question

The value of $\int\limits_0^{2\pi } {\max \left( {\sin x,\cos x} \right)} dx$ is
(a) $3\sqrt 2$
(b) $2\sqrt 2$
(c) 3
(d) None of these

Answer 1

Here, we will use the property of definite integrals to rewrite the given integral as the sum of three integrals. Then we will change the limits for every trigonometric function and simplify the integral. We will then integrate the functions individually, apply the limits and add the terms to find the answer.
Formula Used: The definite integral $\int\limits_a^b {f\left( x \right)} dx$ of a continuous function $f\left( x \right)$ can be written as the sum of the definite integrals $\int\limits_a^c {f\left( x \right)} dx$ and $\int\limits_c^b {f\left( x \right)} dx$, where $c$ lies in the interval $\left( {a,b} \right)$.

Complete step by step solution:
The integral $\int\limits_0^{2\pi } {\max \left( {\sin x,\cos x} \right)} dx$ is a definite integral with the upper limit $2\pi$ and lower limit 0.
We know that the functions $\sin x$ and $\cos x$ are continuous functions.
The definite integral $\int\limits_a^b {f\left( x \right)} dx$ of a continuous function $f\left( x \right)$ can be written as the sum of the definite integrals $\int\limits_a^c {f\left( x \right)} dx$ and $\int\limits_c^b {f\left( x \right)} dx$, where $c$ lies in the interval $\left( {a,b} \right)$.
We will use this property of definite integrals to simplify the given integral.
Therefore, we get
$\int\limits_0^{2\pi } {\max \left( {\sin x,\cos x} \right)} dx = \int\limits_0^{\pi /4} {\max \left( {\sin x,\cos x} \right)} dx + \int\limits_{\pi /4}^{5\pi /4} {\max \left( {\sin x,\cos x} \right)} dx + \int\limits_{5\pi /4}^{2\pi } {\max \left( {\sin x,\cos x} \right)} dx$
Here, $\dfrac{\pi }{4}$ and $\dfrac{{5\pi }}{4}$ lie in the interval $\left( {0,2\pi } \right)$.
Now, we know that $\sin x$ goes from 0 to $\dfrac{1}{{\sqrt 2 }}$, and
$\cos x$ goes from 1 to $\dfrac{1}{{\sqrt 2 }}$ in the interval $\left( {0,\dfrac{\pi }{4}} \right)$.
Thus, $\cos x$ is greater than $\sin x$ in the interval $\left( {0,\dfrac{\pi }{4}} \right)$.
Therefore, the integral
$\int\limits_0^{\pi /4} {\max \left( {\sin x,\cos x} \right)} dx$ becomes $\int\limits_0^{\pi /4} {\cos x} dx$.
Next, we know that $\sin x$ goes from $\dfrac{1}{{\sqrt 2 }}$ to 1, then to 0, and finally to $- \dfrac{1}{{\sqrt 2 }}$ in the interval $\left( {\dfrac{\pi }{4},\dfrac{{5\pi }}{4}} \right)$.
The function $\cos x$ goes from $\dfrac{1}{{\sqrt 2 }}$ to 0, then to $- 1$, and finally back to
$- \dfrac{1}{{\sqrt 2 }}$ in the interval $\left( {\dfrac{\pi }{4},\dfrac{{5\pi }}{4}} \right)$.
Thus, $\sin x$ is greater than $\cos x$ in the interval
$\left( {\dfrac{\pi }{4},\dfrac{{5\pi }}{4}} \right)$.
Therefore, the integral $\int\limits_{\pi /4}^{5\pi /4} {\max \left( {\sin x,\cos x} \right)} dx$ becomes $\int\limits_{\pi /4}^{5\pi /4} {\sin x} dx$.
Finally, we know that $\sin x$ goes from $- \dfrac{1}{{\sqrt 2 }}$ to $- 1$, and then to 0 in the interval $\left( {\dfrac{{5\pi }}{4},2\pi } \right)$.
The function $\cos x$ goes from $- \dfrac{1}{{\sqrt 2 }}$ to 0, and then to 1 in the interval $\left( {\dfrac{{5\pi }}{4},2\pi } \right)$.
Thus, $\cos x$ is greater than
$\sin x$ in the interval $\left( {\dfrac{{5\pi }}{4},2\pi } \right)$.
Therefore, the integral $\int\limits_{5\pi /4}^{2\pi } {\max \left( {\sin x,\cos x} \right)} dx$ becomes $\int\limits_{5\pi /4}^{2\pi } {\cos x} dx$.
Now, we will rewrite the functions in the given integrals.
Therefore, we get
$\Rightarrow \int\limits_0^{2\pi } {\max \left( {\sin x,\cos x} \right)} dx = \int\limits_0^{\pi /4} {\cos x} dx + \int\limits_{\pi /4}^{5\pi /4} {\sin x} dx + \int\limits_{5\pi /4}^{2\pi } {\cos x} dx$
Integrating the functions, we get
$\Rightarrow \int\limits_0^{2\pi } {\max \left( {\sin x,\cos x} \right)} dx = \left. {\left( {\sin x} \right)} \right|_0^{\pi /4} + \left. {\left( { - \cos x} \right)} \right|_{\pi /4}^{5\pi /4} + \left. {\left( {\sin x} \right)} \right|_{5\pi /4}^{2\pi }$
Substituting the limits, we get
$\Rightarrow \int\limits_0^{2\pi } {\max \left( {\sin x,\cos x} \right)} dx = \left( {\sin \dfrac{\pi }{4} - \sin 0} \right) + \left( { - \cos \dfrac{{5\pi }}{4} + \cos \dfrac{\pi }{4}} \right) + \left( {\sin 2\pi - \sin \dfrac{{5\pi }}{4}} \right)$
Substitute the values of the trigonometric ratios, we get
$\begin{array}{l} \Rightarrow \int\limits_0^{2\pi } {\max \left( {\sin x,\cos x} \right)} dx = \left( {\dfrac{1}{{\sqrt 2 }} - 0} \right) + \left[ { - \left( { - \dfrac{1}{{\sqrt 2 }}} \right) + \dfrac{1}{{\sqrt 2 }}} \right] + \left[ {0 - \left( { - \dfrac{1}{{\sqrt 2 }}} \right)} \right]\\\ \Rightarrow \int\limits_0^{2\pi } {\max \left( {\sin x,\cos x} \right)} dx = \dfrac{1}{{\sqrt 2 }} + \dfrac{1}{{\sqrt 2 }} + \dfrac{1}{{\sqrt 2 }} + \dfrac{1}{{\sqrt 2 }}\end{array}$
Adding the terms of the expression, we get
$\begin{array}{l} \Rightarrow \int\limits_0^{2\pi } {\max \left( {\sin x,\cos x} \right)} dx = \dfrac{4}{{\sqrt 2 }}\\\ \Rightarrow \int\limits_0^{2\pi } {\max \left( {\sin x,\cos x} \right)} dx = 2\sqrt 2 \end{array}$
Therefore, we get the value of the integral $\int\limits_0^{2\pi } {\max \left( {\sin x,\cos x} \right)} dx$ as $2\sqrt 2$.

Thus, the correct option is option (b).

Note:
Some common mistakes in this question include using the upper limit in place of the lower limit, and using the property of definite integrals incorrectly. We can also draw a rough graph including the functions $\sin x$ and $\cos x$ to determine the minimum and maximum points.