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What is the value of the integral \[\int\limits_0^\pi {{e^{{{\sin }^2}x}}{{\cos }^3}xdx} \]?
A. \[ - 1\]
B. 0
C. 1
D. \[\pi \]


Answer
VerifiedVerified
162k+ views
Hint: Here, a definite integral is given. First, apply the integration rule \[\int\limits_0^a {f\left( x \right) dx} = \int\limits_0^a {f\left( {a - x} \right) dx} \]. Then, simplify the function by applying the trigonometric formulas \[{\sin ^2}\left( {\pi - x} \right) = {\sin ^2}x\] and \[{\cos ^3}\left( {\pi - x} \right) = - {\cos ^3}x\]. After that, add both integrals and solve them to get the required answer.



Formula Used:\[\int\limits_0^a {f\left( x \right) dx} = \int\limits_0^a {f\left( {a - x} \right) dx} \]
\[{\cos ^3}\left( {\pi - x} \right) = - {\cos ^3}x\]
\[{\sin ^2}\left( {\pi - x} \right) = {\sin ^2}x\]



Complete step by step solution:The given definite integral is \[\int\limits_0^\pi {{e^{{{\sin }^2}x}}{{\cos }^3}xdx} \].

Let consider,
\[I = \int\limits_0^\pi {{e^{{{\sin }^2}x}}{{\cos }^3}xdx} \] \[.....\left( 1 \right)\]
Now apply the property of the definite integral \[\int\limits_0^a {f\left( x \right) dx} = \int\limits_0^a {f\left( {a - x} \right) dx} \].
We get
\[I = \int\limits_0^\pi {{e^{{{\sin }^2}\left( {\pi - x} \right)}}{{\cos }^3}\left( {\pi - x} \right)dx} \]
Apply the trigonometric formulas \[{\cos ^3}\left( {\pi - x} \right) = - {\cos ^3}x\] and \[{\sin ^2}\left( {\pi - x} \right) = {\sin ^2}x\].
\[I = \int\limits_0^\pi {{e^{{{\sin }^2}x}}\left( { - {{\cos }^3}x} \right)dx} \]
\[I = - \int\limits_0^\pi {{e^{{{\sin }^2}x}}{{\cos }^3}xdx} \] \[.....\left( 2 \right)\]
Add the equations \[\left( 1 \right)\] and \[\left( 2 \right)\].
 \[I + I = \int\limits_0^\pi {{e^{{{\sin }^2}x}}{{\cos }^3}xdx} I - \int\limits_0^\pi {{e^{{{\sin }^2}x}}{{\cos }^3}xdx} \]
\[ \Rightarrow 2I = 0\]
\[ \Rightarrow I = 0\]
Thus, \[\int\limits_0^\pi {{e^{{{\sin }^2}x}}{{\cos }^3}xdx} = 0\]



Option ‘B’ is correct


Note: Sometimes students get confused and try to solve the integral by using the trigonometric identities. We can solve this integral by using the methods of indefinite integral. But the answer will be wrong.