Q.18. The value of the integral is:

Q: What is the correct answer for JEE Main Mathematics Q18?

The correct answer is C. Sol. Apply (P-5) property of definite integrals: $\int_a^b f(x) dx = \int_a^b f(a+b-x) dx$. Here, $a = -\pi/4$, $b = \pi/4$. So, $a+b-x = -x$. Let $I = \int_{-\pi/4}^{\pi/4} \frac{32 \cos^4 x}{1 + e^{\sin x}} dx$. Using the property, $I = \int_{-\pi/4}^{\pi/4} \frac{32 \cos^4 (-x)}{1 + e^{\sin (-x)}} dx = \int_{-\pi/4}^{\pi/4} \frac{32 \cos^4 x}{1 + e^{-\sin x}} dx = \int_{-\pi/4}^{\pi/4} \frac{32

Q: Q.18. The value of the integral is:

The correct answer is C. Sol. Apply (P-5) property of definite integrals: $\int_a^b f(x) dx = \int_a^b f(a+b-x) dx$. Here, $a = -\pi/4$, $b = \pi/4$. So, $a+b-x = -x$. Let $I = \int_{-\pi/4}^{\pi/4} \frac{32 \cos^4 x}{1 + e^{\sin x}} dx$. Using the property, $I = \int_{-\pi/4}^{\pi/4} \frac{32 \cos^4 (-x)}{1 + e^{\sin (-x)}} dx = \int_{-\pi/4}^{\pi/4} \frac{32 \cos^4 x}{1 + e^{-\sin x}} dx = \int_{-\pi/4}^{\pi/4} \frac{32

Q: What are the options for this JEE Main Mathematics question?

A. 4\pi + 2 · B. 3\pi + 8 · C. 3\pi + 4 · D. 4\pi + 3

Question

Q.18. The value of the integral  is:

FirstInTest · Accepted Answer

Sol. Apply (P-5) property of definite integrals: $\int_a^b f(x) dx = \int_a^b f(a+b-x) dx$. Here, $a = -\pi/4$, $b = \pi/4$. So, $a+b-x = -x$. Let $I = \int_{-\pi/4}^{\pi/4} \frac{32 \cos^4 x}{1 + e^{\sin x}} dx$. Using the property, $I = \int_{-\pi/4}^{\pi/4} \frac{32 \cos^4 (-x)}{1 + e^{\sin (-x)}} dx = \int_{-\pi/4}^{\pi/4} \frac{32 \cos^4 x}{1 + e^{-\sin x}} dx = \int_{-\pi/4}^{\pi/4} \frac{32 \cos^4 x \cdot e^{\sin x}}{e^{\sin x} + 1} dx$. Adding the two expressions for $I$: $2I = \int_{-\pi/4}^{\pi/4} \frac{32 \cos^4 x (1 + e^{\sin x})}{1 + e^{\sin x}} dx = \int_{-\pi/4}^{\pi/4} 32 \cos^4 x dx$. Since $\cos^4 x$ is an even function, $2I = 2 \int_0^{\pi/4} 32 \cos^4 x dx \Rightarrow I = 32 \int_0^{\pi/4} \cos^4 x dx$. We know $\cos^2 x = \frac{1 + \cos 2x}{2}$. So, $\cos^4 x = (\frac{1 + \cos 2x}{2})^2 = \frac{1}{4} (1 + 2\cos 2x + \cos^2 2x) = \frac{1}{4} (1 + 2\cos 2x + \frac{1 + \cos 4x}{2}) = \frac{1}{8} (2 + 4\cos 2x + 1 + \cos 4x) = \frac{1}{8} (3 + 4\cos 2x + \cos 4x)$. $I = 32 \int_0^{\pi/4} \frac{1}{8} (3 + 4\cos 2x + \cos 4x) dx = 4 [3x + 2\sin 2x + \frac{\sin 4x}{4}]_0^{\pi/4}$. $I = 4 [ (3\frac{\pi}{4} + 2\sin(\frac{\pi}{2}) + \frac{\sin(\pi)}{4}) - (0) ] = 4 [\frac{3\pi}{4} + 2(1) + 0 ] = 3\pi + 8$.

Answer

4\pi + 2

Answer

3\pi + 8

Answer

3\pi + 4

Answer

4\pi + 3

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