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The value of 02πmax(sinx,cosx)dx is
(a) 32
(b) 22
(c) 3
(d) None of these

Answer
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Hint:
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 abf(x)dx of a continuous function f(x) can be written as the sum of the definite integrals acf(x)dx and cbf(x)dx, where c lies in the interval (a,b).

Complete step by step solution:
The integral 02πmax(sinx,cosx)dx is a definite integral with the upper limit 2π and lower limit 0.
We know that the functions sinx and cosx are continuous functions.
The definite integral abf(x)dx of a continuous function f(x) can be written as the sum of the definite integrals acf(x)dx and cbf(x)dx, where c lies in the interval (a,b).
We will use this property of definite integrals to simplify the given integral.
Therefore, we get
02πmax(sinx,cosx)dx=0π/4max(sinx,cosx)dx+π/45π/4max(sinx,cosx)dx+5π/42πmax(sinx,cosx)dx
Here, π4 and 5π4 lie in the interval (0,2π).
Now, we know that sinx goes from 0 to 12, and
cosx goes from 1 to 12 in the interval (0,π4).
Thus, cosx is greater than sinx in the interval (0,π4).
Therefore, the integral
0π/4max(sinx,cosx)dx becomes 0π/4cosxdx.
Next, we know that sinx goes from 12 to 1, then to 0, and finally to 12 in the interval (π4,5π4).
The function cosx goes from 12 to 0, then to 1, and finally back to
12 in the interval (π4,5π4).
Thus, sinx is greater than cosx in the interval
(π4,5π4).
Therefore, the integral π/45π/4max(sinx,cosx)dx becomes π/45π/4sinxdx.
Finally, we know that sinx goes from 12 to 1, and then to 0 in the interval (5π4,2π).
The function cosx goes from 12 to 0, and then to 1 in the interval (5π4,2π).
Thus, cosx is greater than
sinx in the interval (5π4,2π).
Therefore, the integral 5π/42πmax(sinx,cosx)dx becomes 5π/42πcosxdx.
Now, we will rewrite the functions in the given integrals.
Therefore, we get
02πmax(sinx,cosx)dx=0π/4cosxdx+π/45π/4sinxdx+5π/42πcosxdx
Integrating the functions, we get
02πmax(sinx,cosx)dx=(sinx)|0π/4+(cosx)|π/45π/4+(sinx)|5π/42π
Substituting the limits, we get
02πmax(sinx,cosx)dx=(sinπ4sin0)+(cos5π4+cosπ4)+(sin2πsin5π4)
Substitute the values of the trigonometric ratios, we get
02πmax(sinx,cosx)dx=(120)+[(12)+12]+[0(12)]02πmax(sinx,cosx)dx=12+12+12+12
Adding the terms of the expression, we get
02πmax(sinx,cosx)dx=4202πmax(sinx,cosx)dx=22
Therefore, we get the value of the integral 02πmax(sinx,cosx)dx as 22.

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 sinx and cosx to determine the minimum and maximum points.