QJEEM21Maa6be: JEE Main 2021 Mathematics - Class 12 JEEM21Maa6be Solution: The value of the definite integral is equal to :

Question

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The value of the definite integral

\int\limits_{ - {\pi \over 4}}^{{\pi \over 4}} {{{dx} \over {(1 + {e^{x\cos x}})({{\sin }^4}x + {{\cos }^4}x)}}}

is equal to :

Accepted Answer

I = \int\limits_{ - {\pi \over 4}}^{{\pi \over 4}} {{{dx} \over {(1 + {e^{x\cos x}})({{\sin }^4}x + {{\cos }^4}x)}}}

.... (1)

Using

\int\limits_a^b {f(x)dx = \int\limits_a^b {f(a + b - x)dx} }

I = \int\limits_{ - {\pi \over 4}}^{{\pi \over 4}} {{{dx} \over {(1 + {e^{ - x\cos x}})({{\sin }^4}x + {{\cos }^4}x)}}}

Add (1) and (2)

2I = \int\limits_{ - {\pi \over 4}}^{{\pi \over 4}} {{{dx} \over {{{\sin }^4}x + {{\cos }^4}x}}}

2I = 2\int\limits_0^{{\pi \over 4}} {{{dx} \over {{{\sin }^4}x + {{\cos }^4}x}}}

I = \int\limits_0^{{\pi \over 4}} {{{\left( {1 + {1 \over {{{\tan }^2}x}}} \right){{\sec }^2}x} \over {{{\left( {\tan - {1 \over {\tan x}}} \right)}^2} + 2}}} dx

\tan x - {1 \over {\tan x}} = t

\left( {1 + {1 \over {{{\tan }^2}x}}} \right){\sec ^2}xdx = dt

I = \int\limits_{ - \infty }^0 {{{dt} \over {{t^2} + 2}}} = \left[ {{1 \over {\sqrt 2 }}{{\tan }^{ - 1}}\left( {{t \over {\sqrt 2 }}} \right)} \right]_{ - \infty }^0

I = 0 - {1 \over {\sqrt 2 }}\left( { - {\pi \over 2}} \right) = {\pi \over {2\sqrt 2 }}

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