Question

If \[I = \int\limits_0^{100\pi } {\sqrt {\left( {1 - \cos 2x} \right)} } dx\] , then what is the value of \[I\]?
A. \[100\sqrt 2 \]
B. \[200\sqrt 2 \]
C. \[50\sqrt 2 \]
D. None of these

Answer 1

Hint: Here, a definite integral is given. First, simplify the trigonometric term by using the identity \[\cos 2\theta = {\cos ^2}\theta - {\sin ^2}\theta \]. Further simplify it by applying the trigonometric formula \[{\cos ^2}\theta + {\sin ^2}\theta = 1\]. Then, simplify the integral by applying the integration formula \[\int\limits_0^{na} {f\left( x \right)} dx = n\int\limits_0^a {f\left( x \right)} dx\]. After that, solve the integral and apply the limits to get the required answer.



Formula Used:\[\cos 2\theta = {\cos ^2}\theta - {\sin ^2}\theta \]
\[{\cos ^2}\theta + {\sin ^2}\theta = 1\]
\[\int\limits_0^{na} {f\left( x \right)} dx = n\int\limits_0^a {f\left( x \right)} dx\]
\[\int\limits_a^b {\sin xdx = \left[ { - \cos x} \right]} _a^b\]



Complete step by step solution:The given definite integral is \[I = \int\limits_0^{100\pi } {\sqrt {\left( {1 - \cos 2x} \right)} } dx\].
Let’s simplify the integral.
Apply the formula \[\cos 2\theta = {\cos ^2}\theta - {\sin ^2}\theta \]
\[I = \int\limits_0^{100\pi } {\sqrt {\left( {1 - \left( {{{\cos }^2}x - {{\sin }^2}x} \right)} \right)} } dx\]
\[ \Rightarrow I = \int\limits_0^{100\pi } {\sqrt {\left( {1 - {{\cos }^2}x + {{\sin }^2}x} \right)} } dx\]
Apply the trigonometric identity \[{\cos ^2}\theta + {\sin ^2}\theta = 1\].
\[ \Rightarrow I = \int\limits_0^{100\pi } {\sqrt {\left( {{{\sin }^2}x + {{\sin }^2}x} \right)} } dx\]
\[ \Rightarrow I = \int\limits_0^{100\pi } {\sqrt {2{{\sin }^2}x} } dx\]
\[ \Rightarrow I = \int\limits_0^{100\pi } {\sqrt 2 \sin x} dx\]
\[ \Rightarrow I = \sqrt 2 \int\limits_0^{100\pi } {\sin x} dx\]

Now apply the integration rule \[\int\limits_0^{na} {f\left( x \right)} dx = n\int\limits_0^a {f\left( x \right)} dx\].
\[ \Rightarrow I = \sqrt 2 \times 100\int\limits_0^\pi {\sin x} dx\]
Apply the integration formula \[\int\limits_a^b {\sin xdx = \left[ { - \cos x} \right]} _a^b\].
\[ \Rightarrow I = 100\sqrt 2 \left[ { - \cos x} \right]_0^\pi \]
Apply the upper and lower limits.
\[ \Rightarrow I = 100\sqrt 2 \left[ { - \cos \pi - \left( { - \cos 0} \right)} \right]\]
\[ \Rightarrow I = 100\sqrt 2 \left[ { - \left( { - 1} \right) - \left( { - 1} \right)} \right]\]
\[ \Rightarrow I = 100\sqrt 2 \left[ {1 + 1} \right]\]
\[ \Rightarrow I = 100\sqrt 2 \left[ 2 \right]\]
\[ \Rightarrow I = 200\sqrt 2 \]



Option ‘B’ is correct



Note: Students often do mistake to integrating \[\int\limits_a^b {\sin x} dx\] . They apply the formula \[\int\limits_a^b {\sin x} dx = \left[ {\cos x} \right]_a^b\] which is an incorrect formula. They get confused because \[\dfrac{d}{{dx}}\sin x = \cos x\] . The correct formula is \[\int\limits_a^b {\sin x} dx = \left[ { - \cos x} \right]_a^b\].