Question

What is the value of the integral \[\int\limits_0^\pi {{e^{{{\cos }^2}x}}} {\cos ^5}3xdx\]?
A. 1
B. \[ - 1\]
C. 0
D. None of these

Answer 1

Hint: Here, a definite integral is given. First, simplify the integral by using the integral rule \[\int\limits_0^a {f\left( x \right)} dx = \int\limits_0^a {f\left( {a - x} \right)} dx\]. Then, add this integral with the given integral and simplify it. After that, solve the integral using the integration formulas. In the end, apply the upper and lower limits to get the value of the given integral.



Formula Used:\[{\cos ^n}\left( {\pi - \theta } \right) = {\left( { - \cos \theta } \right)^n}\]
Integration rule: \[\int\limits_0^a {f\left( x \right)} dx = \int\limits_0^a {f\left( {a - x} \right)} dx\]
\[\int\limits_a^b n dx = \left[ {nx} \right]_a^b\]



Complete step by step solution:The given definite integral is \[\int\limits_0^\pi {{e^{{{\cos }^2}x}}} {\cos ^5}3xdx\].
Let consider,
\[I = \int\limits_0^\pi {{e^{{{\cos }^2}x}}} {\cos ^5}3xdx\] \[.....\left( 1 \right)\]
Apply the integration rule \[\int\limits_0^a {f\left( x \right)} dx = \int\limits_0^a {f\left( {a - x} \right)} dx\] on the right-hand side.
\[I = \int\limits_0^\pi {{e^{{{\cos }^2}\left( {\pi - x} \right)}}} {\cos ^5}3\left( {\pi - x} \right)dx\]
Apply the exponent property of the trigonometric function \[{\cos ^n}\left( {\pi - \theta } \right) = {\left( { - \cos \theta } \right)^n}\].
\[ \Rightarrow I = \int\limits_0^\pi {{e^{{{\cos }^2}\left( x \right)}}} \left( { - {{\cos }^5}3x} \right)dx\]
\[ \Rightarrow I = - \int\limits_0^\pi {{e^{{{\cos }^2}\left( x \right)}}} {\cos ^5}3xdx\] \[.....\left( 2 \right)\]

Now add the equations \[\left( 1 \right)\] and \[\left( 2 \right)\].
\[I + I = \int\limits_0^\pi {{e^{{{\cos }^2}x}}} {\cos ^5}3xdx - \int\limits_0^\pi {{e^{{{\cos }^2}\left( x \right)}}} {\cos ^5}3xdx\]
\[ \Rightarrow 2I = 0\]
\[ \Rightarrow I = 0\]
Therefore, \[\int\limits_0^\pi {{e^{{{\cos }^2}x}}} {\cos ^5}3xdx = 0\]



Option ‘C’ is correct



Note: Sometimes students get confused about the exponent rule of the trigonometric ratios.
The formulas are as follows:
\[{\sin ^n}\left( {\pi - \theta } \right) = {\sin ^n}\theta \]
\[{\cos ^n}\left( {\pi - \theta } \right) = {\left( { - \cos \theta } \right)^n}\]
\[{\tan ^n}\left( {\pi - \theta } \right) = {\left( { - \tan \theta } \right)^n}\]