<p>If \(a>0\), evaluate \(\displaystyle\int_{-\pi}^{\pi}\frac{\cos^2 x}{1+a^x}\,dx\) [JEE Main 2020]</p>
Step-by-Step Solution
Key Concept: Let I = \intcos^2x/(1+aˣ)dx. King: I = \intcos^2x \cdot aˣ/(1+aˣ)dx. Add: 2I = \intcos^2x dx = \pi \to I = \pi/2.
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<p>Let $I=\int_{-\pi}^\pi\frac{\cos^2 x}{1+a^x}dx$. King ($x\to-x$): $\cos^2(-x)=\cos^2 x$, $a^{-x}=1/a^x$:</p>
<p>$$I=\int_{-\pi}^\pi\frac{\cos^2 x}{1+a^{-x}}dx=\int_{-\pi}^\pi\frac{a^x\cos^2 x}{a^x+1}dx$$</p>
<p>Add: $2I=\int_{-\pi}^\pi\cos^2 x\,dx=\pi\Rightarrow I=\boxed{\dfrac{\pi}{2}}$</p>
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Correct Answer: A