The value of the integral \int\limits_0^\pi / 4 \fracx ~d x\sin ^4(2 x)+\cos ^4(2 x) equals :

Definite Integration PYQ — JEE Main 2024 · 1 Feb · Shift 1 · Q32 with Solution

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Correct answer: C

Let $I=\int_0^{\pi/4} \frac{x\,dx}{\sin^4(2x)+\cos^4(2x)}.$
First simplify the denominator: $\sin^4\theta+\cos^4\theta=(\sin^2\theta+\cos^2\theta)^2-2\sin^2\theta\cos^2\theta=1-2\sin^2\theta\cos^2\theta.$ Since $\sin^2\theta\cos^2\theta=\frac{1}{4}\sin^2 2\theta,$ we get $\sin^4\theta+\cos^4\theta=1-\frac{1}{2}\sin^2 2\theta.$ Now put $\theta=2x$ : $\sin^4(2x)+\cos^4(2x)=1-\frac{1}{2}\sin^2 4x.$ Using $\sin^2 4x=\frac{1-\cos 8x}{2}$ , $1-\frac{1}{2}\sin^2 4x=1-\frac{1}{4}(1-\cos 8x)=\frac{3+\cos 8x}{4}.$ Hence $I=\int_0^{\pi/4} \frac{4x\,dx}{3+\cos 8x}.$
Use the property $\int_0^a x f(x)\,dx=\frac{a}{2}\int_0^a f(x)\,dx$ whenever $f(x)=f(a-x)$ .

Here, with $a=\pi/4$ , $f(x)=\frac{1}{\sin^4(2x)+\cos^4(2x)}.$ Now check symmetry:

=\frac{1}{\cos^4(2x)+\sin^4(2x)}=f(x).$$ So, $$I=\frac{\pi}{8}\int_0^{\pi/4}\frac{dx}{\sin^4(2x)+\cos^4(2x)}.$$ 4. Let $$J=\int_0^{\pi/4}\frac{dx}{\sin^4(2x)+\cos^4(2x)}.$$ Then using the simplified form, $$J=\int_0^{\pi/4}\frac{4\,dx}{3+\cos 8x}.$$ Put $t=8x$, so $dt=8dx$, $dx=dt/8$. Limits: $x=0\to t=0$, $x=\pi/4\to t=2\pi$. Thus, $$J=4\int_0^{\pi/4}\frac{dx}{3+\cos 8x} =4\cdot \frac{1}{8}\int_0^{2\pi}\frac{dt}{3+\cos t} =\frac{1}{2}\int_0^{2\pi}\frac{dt}{3+\cos t}.$$ 5. Use the standard result $$\int_0^{2\pi}\frac{dt}{a+b\cos t}=\frac{2\pi}{\sqrt{a^2-b^2}},\qquad a>|b|.$$ Here $a=3$, $b=1$, so $$\int_0^{2\pi}\frac{dt}{3+\cos t}=\frac{2\pi}{\sqrt{9-1}}=\frac{2\pi}{\sqrt{8}}=\frac{\pi}{\sqrt{2}}.$$ Therefore, $$J=\frac{1}{2}\cdot \frac{\pi}{\sqrt{2}}=\frac{\pi}{2\sqrt{2}}.$$ 6. Now $$I=\frac{\pi}{8}J=\frac{\pi}{8}\cdot \frac{\pi}{2\sqrt{2}}=\frac{\pi^2}{16\sqrt{2}}=\frac{\sqrt{2}\pi^2}{32}.$$ 7. Hence the correct option is $$\boxed{\text{C }\frac{\sqrt{2}\pi^2}{32}}.$$

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Correct answer: C

Let $I=\int_0^{\pi/4} \frac{x\,dx}{\sin^4(2x)+\cos^4(2x)}.$
First simplify the denominator: $\sin^4\theta+\cos^4\theta=(\sin^2\theta+\cos^2\theta)^2-2\sin^2\theta\cos^2\theta=1-2\sin^2\theta\cos^2\theta.$ Since $\sin^2\theta\cos^2\theta=\frac{1}{4}\sin^2 2\theta,$ we get $\sin^4\theta+\cos^4\theta=1-\frac{1}{2}\sin^2 2\theta.$ Now put $\theta=2x$ : $\sin^4(2x)+\cos^4(2x)=1-\frac{1}{2}\sin^2 4x.$ Using $\sin^2 4x=\frac{1-\cos 8x}{2}$ , $1-\frac{1}{2}\sin^2 4x=1-\frac{1}{4}(1-\cos 8x)=\frac{3+\cos 8x}{4}.$ Hence $I=\int_0^{\pi/4} \frac{4x\,dx}{3+\cos 8x}.$
Use the property $\int_0^a x f(x)\,dx=\frac{a}{2}\int_0^a f(x)\,dx$ whenever $f(x)=f(a-x)$ .

Here, with $a=\pi/4$ , $f(x)=\frac{1}{\sin^4(2x)+\cos^4(2x)}.$ Now check symmetry:

=\frac{1}{\cos^4(2x)+\sin^4(2x)}=f(x).$$ So, $$I=\frac{\pi}{8}\int_0^{\pi/4}\frac{dx}{\sin^4(2x)+\cos^4(2x)}.$$ 4. Let $$J=\int_0^{\pi/4}\frac{dx}{\sin^4(2x)+\cos^4(2x)}.$$ Then using the simplified form, $$J=\int_0^{\pi/4}\frac{4\,dx}{3+\cos 8x}.$$ Put $t=8x$, so $dt=8dx$, $dx=dt/8$. Limits: $x=0\to t=0$, $x=\pi/4\to t=2\pi$. Thus, $$J=4\int_0^{\pi/4}\frac{dx}{3+\cos 8x} =4\cdot \frac{1}{8}\int_0^{2\pi}\frac{dt}{3+\cos t} =\frac{1}{2}\int_0^{2\pi}\frac{dt}{3+\cos t}.$$ 5. Use the standard result $$\int_0^{2\pi}\frac{dt}{a+b\cos t}=\frac{2\pi}{\sqrt{a^2-b^2}},\qquad a>|b|.$$ Here $a=3$, $b=1$, so $$\int_0^{2\pi}\frac{dt}{3+\cos t}=\frac{2\pi}{\sqrt{9-1}}=\frac{2\pi}{\sqrt{8}}=\frac{\pi}{\sqrt{2}}.$$ Therefore, $$J=\frac{1}{2}\cdot \frac{\pi}{\sqrt{2}}=\frac{\pi}{2\sqrt{2}}.$$ 6. Now $$I=\frac{\pi}{8}J=\frac{\pi}{8}\cdot \frac{\pi}{2\sqrt{2}}=\frac{\pi^2}{16\sqrt{2}}=\frac{\sqrt{2}\pi^2}{32}.$$ 7. Hence the correct option is $$\boxed{\text{C }\frac{\sqrt{2}\pi^2}{32}}.$$