Question

The value of the definite integral $\int_{-\pi/4}^{\pi/4} \frac{dx}{(1+e^{x\cos x})(\sin^4 x + \cos^4 x)}$ is equal to :

• A. $\frac{\pi}{2\sqrt{2}}$
• B. $-\frac{\pi}{4}$
• C. $-\frac{\pi}{2}$
• D. $\frac{\pi}{\sqrt{2}}$

Hint:

In definite integrals with limits $[-a,a]$, always check for symmetry: \[ \int_{-a}^{a}f(x)\,dx=\int_{0}^{a}[f(x)+f(-x)]\,dx \] This trick is extremely powerful when exponential terms like $e^x$ or $e^{x\cos x}$ are involved.

Answer 1

The Correct Option is A

Let \[ I=\int_{-\pi/4}^{\pi/4}\frac{dx}{(1+e^{x\cos x})(\sin^4x+\cos^4x)} \] Step 1: Use symmetry of limits
For any function $f(x)$, \[ \int_{-a}^{a} f(x)\,dx=\int_{0}^{a}[f(x)+f(-x)]\,dx \] Define \[ f(x)=\frac{1}{(1+e^{x\cos x})(\sin^4x+\cos^4x)} \] Using $\cos(-x)=\cos x$ and $\sin(-x)=-\sin x$: \[ f(-x)=\frac{1}{(1+e^{-x\cos x})(\sin^4x+\cos^4x)} \] Step 2: Add $f(x)$ and $f(-x)$
\[ f(x)+f(-x) =\frac{1}{\sin^4x+\cos^4x} \left(\frac{1}{1+e^{x\cos x}}+\frac{1}{1+e^{-x\cos x}}\right) \] Let $u=e^{x\cos x}$: \[ \frac{1}{1+u}+\frac{1}{1+1/u} =\frac{1}{1+u}+\frac{u}{1+u}=1 \] Hence, \[ f(x)+f(-x)=\frac{1}{\sin^4x+\cos^4x} \] So, \[ I=\int_{0}^{\pi/4}\frac{dx}{\sin^4x+\cos^4x} \] Step 3: Simplify the denominator
\[ \sin^4x+\cos^4x =(\sin^2x+\cos^2x)^2-2\sin^2x\cos^2x =1-\tfrac12\sin^22x \] Thus, \[ I=\int_{0}^{\pi/4}\frac{dx}{1-\tfrac12\sin^22x} \] Let $u=2x$, $du=2dx$: \[ I=\frac12\int_{0}^{\pi/2}\frac{du}{1-\tfrac12\sin^2u} =\frac12\int_{0}^{\pi/2}\frac{2\,du}{2-\sin^2u} \] Step 4: Use standard integral
\[ \int_{0}^{\pi/2}\frac{du}{a+b\sin^2u} =\frac{\pi}{2\sqrt{a(a+b)}} \] Here, $a=2$, $b=-1$: \[ I=\frac{\pi}{2\sqrt{2}} \] \[ \boxed{\text{Correct Answer: (A) }\dfrac{\pi}{2\sqrt{2}}} \]