GATE 2025 IN Q.38: solution and answer

Official answer: CSource: Official answer key, GATE 2025 IN (IIT Roorkee)

1. Since $\cos (-x)=\cos x$, the integrand $f(x)={\cos }^{6}x+{\cos }^{4}x$ is even, so ${\int }_{-\pi }^{\pi }f(x)dx=2{\int }_0^{\pi }f(x)dx$.

2. For even $n$, ${\cos }^n(\pi -x)=(-\cos x{)}^n={\cos }^{n}x$. Substituting $x\to \pi -x$ on $[\pi /2,\pi ]$ maps it onto $[0,\pi /2]$, so ${\int }_0^{\pi }{\cos }^{n}xdx=2{\int }_0^{\pi /2}{\cos }^{n}xdx$. Combined with step 1: ${\int }_{-\pi }^{\pi }{\cos }^{n}xdx=4{\int }_0^{\pi /2}{\cos }^{n}xdx$ for even $n$.

3. Wallis' formula for even $n$: ${\int }_0^{\pi /2}{\cos }^{n}xdx=\frac{(n-1)(n-3)\cdots 1}{n(n-2)\cdots 2}\cdot \frac{\pi }{2}$.

4. For $n=6$: ${\int }_0^{\pi /2}{\cos }^{6}xdx=\frac{5\cdot 3\cdot 1}{6\cdot 4\cdot 2}\cdot \frac{\pi }{2}=\frac{5\pi }{32}$, so ${\int }_{-\pi }^{\pi }{\cos }^{6}xdx=4\cdot \frac{5\pi }{32}=\frac{5\pi }{8}$.

5. For $n=4$: ${\int }_0^{\pi /2}{\cos }^{4}xdx=\frac{3\cdot 1}{4\cdot 2}\cdot \frac{\pi }{2}=\frac{3\pi }{16}$, so ${\int }_{-\pi }^{\pi }{\cos }^{4}xdx=4\cdot \frac{3\pi }{16}=\frac{3\pi }{4}=\frac{6\pi }{8}$.

6. Add the two: $${\int }_{-\pi }^{\pi }\left({\cos }^{6}x+{\cos }^{4}x\right)dx=\frac{5\pi }{8}+\frac{6\pi }{8}=\frac{11\pi }{8}.$$ This is option (C).