If I = 2 \over \pi \int\limits_ - \pi /4^\pi /4 dx \over (1 + e^\sin x)(2 - \cos 2x) , then 27I2 equals .................

Definite Integration PYQ — JEE Advanced 2019 · Shift 1 · Q34 with Solution

View written solutionFree

Correct answer: 4

Step-by-step derivation:

Step 1: Simplify the integral using properties of definite integrals.

The given integral is:

over \pi }\int\limits_{ - \pi /4}^{\pi /4} {{{dx} \over {(1 + {e^{\sin x}})(2 - \cos 2x)}}}$$ The integral is of the form $\int_{-a}^a f(x) dx$. We can use the property: $$ \int_{-a}^a f(x) dx = \int_0^a (f(x) + f(-x)) dx $$ Let the integrand be $g(x) = {1 \over {(1 + {e^{\sin x}})(2 - \cos 2x)}}$. Now, let's find $g(-x)$: $$ g(-x) = {1 \over {(1 + e^{\sin(-x)})(2 - \cos(2(-x)))}} $$ Since $\sin(-x) = -\sin x$ and $\cos(-2x) = \cos(2x)$, we have: $$ g(-x) = {1 \over {(1 + e^{-\sin x})(2 - \cos 2x)}} = {1 \over {(1 + {1 \over e^{\sin x}})(2 - \cos 2x)}} = {e^{\sin x} \over {(e^{\sin x} + 1)(2 - \cos 2x)}} $$ Now, let's compute the sum $g(x) + g(-x)$: $$ g(x) + g(-x) = {1 \over {(1 + e^{\sin x})(2 - \cos 2x)}} + {e^{\sin x} \over {(1 + e^{\sin x})(2 - \cos 2x)}} $$ $$ g(x) + g(-x) = {1 + e^{\sin x} \over {(1 + e^{\sin x})(2 - \cos 2x)}} = {1 \over {2 - \cos 2x}} $$ Applying the property to the integral $I$: $$ I = {2 \over \pi } \int_0^{\pi/4} (g(x) + g(-x)) dx $$ $$ I = {2 \over \pi } \int_0^{\pi/4} {1 \over {2 - \cos 2x}} dx $$ **Step 2: Evaluate the simplified integral.** To evaluate the integral, we use the half-angle formula for cosine in terms of tangent: $\cos 2x = {{1 - \tan^2 x} \over {1 + \tan^2 x}}$. $$ I = {2 \over \pi } \int_0^{\pi/4} {1 \over {2 - \frac{1 - \tan^2 x}{1 + \tan^2 x}}} dx $$ $$ I = {2 \over \pi } \int_0^{\pi/4} {1 + \tan^2 x \over {2(1 + \tan^2 x) - (1 - \tan^2 x)}} dx $$ $$ I = {2 \over \pi } \int_0^{\pi/4} {\sec^2 x \over {2 + 2\tan^2 x - 1 + \tan^2 x}} dx $$ $$ I = {2 \over \pi } \int_0^{\pi/4} {\sec^2 x \over {1 + 3\tan^2 x}} dx $$ Now, we use the substitution $t = \tan x$. This implies $dt = \sec^2 x dx$. The limits of integration also change: - When $x=0$, $t = \tan(0) = 0$. - When $x=\pi/4$, $t = \tan(\pi/4) = 1$. The integral becomes: $$ I = {2 \over \pi } \int_0^1 {{dt} \over {1 + 3t^2}} = {2 \over \pi } \int_0^1 {{dt} \over {1 + (\sqrt{3}t)^2}} $$ This is a standard integral of the form $\int \frac{du}{a^2+u^2} = \frac{1}{a}\arctan(\frac{u}{a})$. Here $a=1$ and $u=\sqrt{3}t$. $$ I = {2 \over \pi } \left[ {1 \over \sqrt{3}} \arctan(\sqrt{3}t) \right]_0^1 $$ $$ I = {2 \over {\pi \sqrt{3}}} [\arctan(\sqrt{3}t)]_0^1 $$ $$ I = {2 \over {\pi \sqrt{3}}} (\arctan(\sqrt{3}) - \arctan(0)) $$ $$ I = {2 \over {\pi \sqrt{3}}} \left( {\pi \over 3} - 0 \right) $$ $$ I = {2 \over {\pi \sqrt{3}}} \cdot {\pi \over 3} = {2 \over {3\sqrt{3}}} $$ **Step 3: Calculate the final value.** We need to find the value of $27I^2$. First, calculate $I^2$: $$ I^2 = \left( {2 \over {3\sqrt{3}}} \right)^2 = {4 \over {9 \times 3}} = {4 \over 27} $$ Now, multiply by 27: $$ 27I^2 = 27 \times \left( {4 \over 27} \right) = 4 $$ Thus, the value of $27I^2$ is 4.

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View written solutionFree

Correct answer: 4

Step-by-step derivation:

Step 1: Simplify the integral using properties of definite integrals.

The given integral is:

over \pi }\int\limits_{ - \pi /4}^{\pi /4} {{{dx} \over {(1 + {e^{\sin x}})(2 - \cos 2x)}}}$$ The integral is of the form $\int_{-a}^a f(x) dx$. We can use the property: $$ \int_{-a}^a f(x) dx = \int_0^a (f(x) + f(-x)) dx $$ Let the integrand be $g(x) = {1 \over {(1 + {e^{\sin x}})(2 - \cos 2x)}}$. Now, let's find $g(-x)$: $$ g(-x) = {1 \over {(1 + e^{\sin(-x)})(2 - \cos(2(-x)))}} $$ Since $\sin(-x) = -\sin x$ and $\cos(-2x) = \cos(2x)$, we have: $$ g(-x) = {1 \over {(1 + e^{-\sin x})(2 - \cos 2x)}} = {1 \over {(1 + {1 \over e^{\sin x}})(2 - \cos 2x)}} = {e^{\sin x} \over {(e^{\sin x} + 1)(2 - \cos 2x)}} $$ Now, let's compute the sum $g(x) + g(-x)$: $$ g(x) + g(-x) = {1 \over {(1 + e^{\sin x})(2 - \cos 2x)}} + {e^{\sin x} \over {(1 + e^{\sin x})(2 - \cos 2x)}} $$ $$ g(x) + g(-x) = {1 + e^{\sin x} \over {(1 + e^{\sin x})(2 - \cos 2x)}} = {1 \over {2 - \cos 2x}} $$ Applying the property to the integral $I$: $$ I = {2 \over \pi } \int_0^{\pi/4} (g(x) + g(-x)) dx $$ $$ I = {2 \over \pi } \int_0^{\pi/4} {1 \over {2 - \cos 2x}} dx $$ **Step 2: Evaluate the simplified integral.** To evaluate the integral, we use the half-angle formula for cosine in terms of tangent: $\cos 2x = {{1 - \tan^2 x} \over {1 + \tan^2 x}}$. $$ I = {2 \over \pi } \int_0^{\pi/4} {1 \over {2 - \frac{1 - \tan^2 x}{1 + \tan^2 x}}} dx $$ $$ I = {2 \over \pi } \int_0^{\pi/4} {1 + \tan^2 x \over {2(1 + \tan^2 x) - (1 - \tan^2 x)}} dx $$ $$ I = {2 \over \pi } \int_0^{\pi/4} {\sec^2 x \over {2 + 2\tan^2 x - 1 + \tan^2 x}} dx $$ $$ I = {2 \over \pi } \int_0^{\pi/4} {\sec^2 x \over {1 + 3\tan^2 x}} dx $$ Now, we use the substitution $t = \tan x$. This implies $dt = \sec^2 x dx$. The limits of integration also change: - When $x=0$, $t = \tan(0) = 0$. - When $x=\pi/4$, $t = \tan(\pi/4) = 1$. The integral becomes: $$ I = {2 \over \pi } \int_0^1 {{dt} \over {1 + 3t^2}} = {2 \over \pi } \int_0^1 {{dt} \over {1 + (\sqrt{3}t)^2}} $$ This is a standard integral of the form $\int \frac{du}{a^2+u^2} = \frac{1}{a}\arctan(\frac{u}{a})$. Here $a=1$ and $u=\sqrt{3}t$. $$ I = {2 \over \pi } \left[ {1 \over \sqrt{3}} \arctan(\sqrt{3}t) \right]_0^1 $$ $$ I = {2 \over {\pi \sqrt{3}}} [\arctan(\sqrt{3}t)]_0^1 $$ $$ I = {2 \over {\pi \sqrt{3}}} (\arctan(\sqrt{3}) - \arctan(0)) $$ $$ I = {2 \over {\pi \sqrt{3}}} \left( {\pi \over 3} - 0 \right) $$ $$ I = {2 \over {\pi \sqrt{3}}} \cdot {\pi \over 3} = {2 \over {3\sqrt{3}}} $$ **Step 3: Calculate the final value.** We need to find the value of $27I^2$. First, calculate $I^2$: $$ I^2 = \left( {2 \over {3\sqrt{3}}} \right)^2 = {4 \over {9 \times 3}} = {4 \over 27} $$ Now, multiply by 27: $$ 27I^2 = 27 \times \left( {4 \over 27} \right) = 4 $$ Thus, the value of $27I^2$ is 4.

Definite Integration question

Definite Integration question

Step-by-step derivation:

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Definite Integration question

Definite Integration question

Step-by-step derivation:

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