
Solve $\int_{0}^{\pi/2}\surd\cos~x/\surd\sin~x +\surd\cos~x$ [MNR 1989; UPSEAT 2002]
(A) 0
(B) $\\pi/2$
(C) $\\pi/4$
(D) None of these
Answer
161.4k+ views
Hint: Integration is part of calculus. The process of combining smaller parts into a single, integrated system is known as integration. It is the reverse of differentiation. It is also called anti differentiation.
Formula Used:The fundamental property of integration $\int_{0}^{a}f(x)=\int_{0}^{a}f(a-x)$
is used to solve this question. The integration is solved, and then the upper and lower limits are substituted.
Complete step by step solution:The given integral is: I = $\int_{0}^{\pi/2}\surd\cos~x/\surd\sin~x +\surd\cos~x$ ---------- (1)
Here the given function is f(x) = $\\surd\cos~x/\surd\sin~x +\surd\cos~x$ and the total range over which the integral is to be done is 0 to $\\pi/2$, with a lower limit of 0 and an upper limit of $\\pi/2$.
We know that the fundamental property of integrals is $\int_{0}^{a}f(x)=\int_{0}^{a}f(a-x)$.
Using this property, we get
I = $\int_{0}^{\pi/2}\surd\cos~(\pi/2-x)/\surd\sin~(\pi/2-x) +\surd\cos~(\pi/2-x)$
We know that $\\sin(\pi/2-x)=cos~x$ and $\\cos(\pi/2-x)=sin~x$
I = $\int_{0}^{\pi/2}\surd\sin~x/\surd\cos~x +\surd\sin~x$ ------- (2)
Adding equations (1) and (2), we get
2I = $\int_{0}^{\pi/2}\surd\cos~x/\surd\sin~x +\surd\cos~x+\int_{0}^{\pi/2}\surd\sin~x/\surd\sin~x +\surd\cos~x$
2I = $\int_{0}^{\pi/2}\surd\cos~x +\surd\sin~x/\surd\sin~x +\surd\cos~x$
2I = $\int_{0}^{\pi/2}1.{\text{d}x}$
2I = $\left[x\right]_{0}^{\pi/2}$
2I =$\ (\dfrac{\pi}{2}-0)$
2I = $\dfrac{\pi}{2}$
I = $\dfrac{\pi}{4}$
Hence the integration of $\\surd\cos~x/\surd\sin~x +\surd\cos~x$ over the range 0 to $\dfrac{\pi}{2}$ is $\dfrac{\pi}{4}$ .
Option ‘C’ is correct
Note: The integration should be done carefully. We should not forget to apply the limits to the integral. Integrals with no integration limit are known as indefinite integrals. Integrals with an upper and lower limit are said to be definite integrals. Integration has various applications in daily life. It is also used to calculate the centre of mass, centre of gravity, mass, and momentum of the satellites. It is used to calculate the area under the curves.
Formula Used:The fundamental property of integration $\int_{0}^{a}f(x)=\int_{0}^{a}f(a-x)$
is used to solve this question. The integration is solved, and then the upper and lower limits are substituted.
Complete step by step solution:The given integral is: I = $\int_{0}^{\pi/2}\surd\cos~x/\surd\sin~x +\surd\cos~x$ ---------- (1)
Here the given function is f(x) = $\\surd\cos~x/\surd\sin~x +\surd\cos~x$ and the total range over which the integral is to be done is 0 to $\\pi/2$, with a lower limit of 0 and an upper limit of $\\pi/2$.
We know that the fundamental property of integrals is $\int_{0}^{a}f(x)=\int_{0}^{a}f(a-x)$.
Using this property, we get
I = $\int_{0}^{\pi/2}\surd\cos~(\pi/2-x)/\surd\sin~(\pi/2-x) +\surd\cos~(\pi/2-x)$
We know that $\\sin(\pi/2-x)=cos~x$ and $\\cos(\pi/2-x)=sin~x$
I = $\int_{0}^{\pi/2}\surd\sin~x/\surd\cos~x +\surd\sin~x$ ------- (2)
Adding equations (1) and (2), we get
2I = $\int_{0}^{\pi/2}\surd\cos~x/\surd\sin~x +\surd\cos~x+\int_{0}^{\pi/2}\surd\sin~x/\surd\sin~x +\surd\cos~x$
2I = $\int_{0}^{\pi/2}\surd\cos~x +\surd\sin~x/\surd\sin~x +\surd\cos~x$
2I = $\int_{0}^{\pi/2}1.{\text{d}x}$
2I = $\left[x\right]_{0}^{\pi/2}$
2I =$\ (\dfrac{\pi}{2}-0)$
2I = $\dfrac{\pi}{2}$
I = $\dfrac{\pi}{4}$
Hence the integration of $\\surd\cos~x/\surd\sin~x +\surd\cos~x$ over the range 0 to $\dfrac{\pi}{2}$ is $\dfrac{\pi}{4}$ .
Option ‘C’ is correct
Note: The integration should be done carefully. We should not forget to apply the limits to the integral. Integrals with no integration limit are known as indefinite integrals. Integrals with an upper and lower limit are said to be definite integrals. Integration has various applications in daily life. It is also used to calculate the centre of mass, centre of gravity, mass, and momentum of the satellites. It is used to calculate the area under the curves.
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