The value of the definite integral \int\limits_ - \pi \over 4^\pi \over 4 dx \over (1 + e^x\cos x)(\sin ^4x + \cos ^4x) is equal to :

Definite Integration PYQ — JEE Main 2021 · 27 Jul · Shift 1 · Q25 with Solution

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Correct answer: B

Let $I=\int_{-\pi/4}^{\pi/4}\frac{dx}{\left(1+e^{x\cos x}\right)\left(\sin^4 x+\cos^4 x\right)}.$ We will use the standard symmetry property on a symmetric interval: $\int_{-a}^{a} f(x)\,dx=\int_{-a}^{a} f(-x)\,dx,$ and hence $2I=\int_{-a}^{a}\bigl(f(x)+f(-x)\bigr)\,dx.$ Here, $f(x)=\frac{1}{\left(1+e^{x\cos x}\right)\left(\sin^4 x+\cos^4 x\right)}.$
Compute $f(-x)$ . Since $\cos(-x)=\cos x$ , $\sin(-x)=-\sin x$ , we get $(-x)\cos(-x)=-x\cos x,$ and $\sin^4(-x)+\cos^4(-x)=\sin^4 x+\cos^4 x.$ Therefore, $f(-x)=\frac{1}{\left(1+e^{-x\cos x}\right)\left(\sin^4 x+\cos^4 x\right)}.$
Add $f(x)$ and $f(-x)$ : $f(x)+f(-x)=\frac{1}{\sin^4 x+\cos^4 x}\left(\frac{1}{1+e^{x\cos x}}+\frac{1}{1+e^{-x\cos x}}\right).$ Now use $\frac{1}{1+e^t}+\frac{1}{1+e^{-t}}=1.$ So, $f(x)+f(-x)=\frac{1}{\sin^4 x+\cos^4 x}.$ Hence, $2I=\int_{-\pi/4}^{\pi/4}\frac{dx}{\sin^4 x+\cos^4 x}.$ Thus $I=\frac12\int_{-\pi/4}^{\pi/4}\frac{dx}{\sin^4 x+\cos^4 x}.$
Simplify the trigonometric denominator. Using $\sin^4 x+\cos^4 x=(\sin^2 x+\cos^2 x)^2-2\sin^2 x\cos^2 x=1-\frac12\sin^2 2x.$ Also, $\sin^4 x+\cos^4 x=\frac{1+\cos^2 2x}{2}.$ Therefore, $\frac{1}{\sin^4 x+\cos^4 x}=\frac{2}{1+\cos^2 2x}.$ So,

\int_{-\pi/4}^{\pi/4}\frac{dx}{1+\cos^2 2x}.$$ 5. Substitute $t=\tan x$. Then $$dx=\frac{dt}{1+t^2},$$ and as $x$ goes from $-\pi/4$ to $\pi/4$, $t$ goes from $-1$ to $1$. Also, $$\cos 2x=\frac{1-t^2}{1+t^2}.$$ So $$1+\cos^2 2x=1+\left(\frac{1-t^2}{1+t^2}\right)^2 =\frac{(1+t^2)^2+(1-t^2)^2}{(1+t^2)^2} =\frac{2(1+t^4)}{(1+t^2)^2}.$$ Hence, $$\frac{dx}{1+\cos^2 2x} =\frac{1}{1+\cos^2 2x}\cdot \frac{dt}{1+t^2} =\frac{(1+t^2)^2}{2(1+t^4)}\cdot\frac{dt}{1+t^2} =\frac{1+t^2}{2(1+t^4)}dt.$$ Therefore, $$I=\int_{-1}^{1}\frac{1+t^2}{2(1+t^4)}dt.$$ 6. Factor $t^4+1$: $$t^4+1=(t^2+\sqrt2 t+1)(t^2-\sqrt2 t+1).$$ A useful identity is $$\frac{1+t^2}{1+t^4}=\frac12\left(\frac{1}{t^2+\sqrt2 t+1}+\frac{1}{t^2-\sqrt2 t+1}\right)\cdot 2,$$ more directly, $$\frac{1+t^2}{1+t^4}=\frac12\left(\frac{1}{t^2+\sqrt2 t+1}+\frac{1}{t^2-\sqrt2 t+1}\right)\cdot 2,$$ which simplifies to $$\frac{1+t^2}{2(1+t^4)}=\frac14\left(\frac{1}{t^2+\sqrt2 t+1}+\frac{1}{t^2-\sqrt2 t+1}\right).$$ Thus, $$I=\frac14\int_{-1}^{1}\left(\frac{1}{(t+1/\sqrt2)^2+1/2}+\frac{1}{(t-1/\sqrt2)^2+1/2}\right)dt.$$ Now integrate using $$\int \frac{du}{u^2+a^2}=\frac1a\tan^{-1}\left(\frac{u}{a}\right).$$ Since $a=1/\sqrt2$, $$I=\frac{\sqrt2}{4}\left[\tan^{-1}(\sqrt2 t+1)+\tan^{-1}(\sqrt2 t-1)\right]_{-1}^{1}.$$ 7. Evaluate at the limits. At $t=1$: $$\tan^{-1}(\sqrt2+1)+\tan^{-1}(\sqrt2-1).$$ Using $$(\sqrt2+1)(\sqrt2-1)=1,$$ both positive, so $$\tan^{-1}(\sqrt2+1)+\tan^{-1}(\sqrt2-1)=\frac\pi2.$$ At $t=-1$: $$\tan^{-1}(1-\sqrt2)+\tan^{-1}(-(\sqrt2+1))=-\frac\pi2.$$ Therefore the total bracket value is $$\frac\pi2-\left(-\frac\pi2\right)=\pi.$$ So, $$I=\frac{\sqrt2}{4}\cdot \pi=\frac{\pi}{2\sqrt2}.$$ 8. Hence the value of the integral is $$\boxed{\frac{\pi}{2\sqrt2}}.$$ This corresponds to **Option B**.

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Correct answer: B

Let $I=\int_{-\pi/4}^{\pi/4}\frac{dx}{\left(1+e^{x\cos x}\right)\left(\sin^4 x+\cos^4 x\right)}.$ We will use the standard symmetry property on a symmetric interval: $\int_{-a}^{a} f(x)\,dx=\int_{-a}^{a} f(-x)\,dx,$ and hence $2I=\int_{-a}^{a}\bigl(f(x)+f(-x)\bigr)\,dx.$ Here, $f(x)=\frac{1}{\left(1+e^{x\cos x}\right)\left(\sin^4 x+\cos^4 x\right)}.$
Compute $f(-x)$ . Since $\cos(-x)=\cos x$ , $\sin(-x)=-\sin x$ , we get $(-x)\cos(-x)=-x\cos x,$ and $\sin^4(-x)+\cos^4(-x)=\sin^4 x+\cos^4 x.$ Therefore, $f(-x)=\frac{1}{\left(1+e^{-x\cos x}\right)\left(\sin^4 x+\cos^4 x\right)}.$
Add $f(x)$ and $f(-x)$ : $f(x)+f(-x)=\frac{1}{\sin^4 x+\cos^4 x}\left(\frac{1}{1+e^{x\cos x}}+\frac{1}{1+e^{-x\cos x}}\right).$ Now use $\frac{1}{1+e^t}+\frac{1}{1+e^{-t}}=1.$ So, $f(x)+f(-x)=\frac{1}{\sin^4 x+\cos^4 x}.$ Hence, $2I=\int_{-\pi/4}^{\pi/4}\frac{dx}{\sin^4 x+\cos^4 x}.$ Thus $I=\frac12\int_{-\pi/4}^{\pi/4}\frac{dx}{\sin^4 x+\cos^4 x}.$
Simplify the trigonometric denominator. Using $\sin^4 x+\cos^4 x=(\sin^2 x+\cos^2 x)^2-2\sin^2 x\cos^2 x=1-\frac12\sin^2 2x.$ Also, $\sin^4 x+\cos^4 x=\frac{1+\cos^2 2x}{2}.$ Therefore, $\frac{1}{\sin^4 x+\cos^4 x}=\frac{2}{1+\cos^2 2x}.$ So,

\int_{-\pi/4}^{\pi/4}\frac{dx}{1+\cos^2 2x}.$$ 5. Substitute $t=\tan x$. Then $$dx=\frac{dt}{1+t^2},$$ and as $x$ goes from $-\pi/4$ to $\pi/4$, $t$ goes from $-1$ to $1$. Also, $$\cos 2x=\frac{1-t^2}{1+t^2}.$$ So $$1+\cos^2 2x=1+\left(\frac{1-t^2}{1+t^2}\right)^2 =\frac{(1+t^2)^2+(1-t^2)^2}{(1+t^2)^2} =\frac{2(1+t^4)}{(1+t^2)^2}.$$ Hence, $$\frac{dx}{1+\cos^2 2x} =\frac{1}{1+\cos^2 2x}\cdot \frac{dt}{1+t^2} =\frac{(1+t^2)^2}{2(1+t^4)}\cdot\frac{dt}{1+t^2} =\frac{1+t^2}{2(1+t^4)}dt.$$ Therefore, $$I=\int_{-1}^{1}\frac{1+t^2}{2(1+t^4)}dt.$$ 6. Factor $t^4+1$: $$t^4+1=(t^2+\sqrt2 t+1)(t^2-\sqrt2 t+1).$$ A useful identity is $$\frac{1+t^2}{1+t^4}=\frac12\left(\frac{1}{t^2+\sqrt2 t+1}+\frac{1}{t^2-\sqrt2 t+1}\right)\cdot 2,$$ more directly, $$\frac{1+t^2}{1+t^4}=\frac12\left(\frac{1}{t^2+\sqrt2 t+1}+\frac{1}{t^2-\sqrt2 t+1}\right)\cdot 2,$$ which simplifies to $$\frac{1+t^2}{2(1+t^4)}=\frac14\left(\frac{1}{t^2+\sqrt2 t+1}+\frac{1}{t^2-\sqrt2 t+1}\right).$$ Thus, $$I=\frac14\int_{-1}^{1}\left(\frac{1}{(t+1/\sqrt2)^2+1/2}+\frac{1}{(t-1/\sqrt2)^2+1/2}\right)dt.$$ Now integrate using $$\int \frac{du}{u^2+a^2}=\frac1a\tan^{-1}\left(\frac{u}{a}\right).$$ Since $a=1/\sqrt2$, $$I=\frac{\sqrt2}{4}\left[\tan^{-1}(\sqrt2 t+1)+\tan^{-1}(\sqrt2 t-1)\right]_{-1}^{1}.$$ 7. Evaluate at the limits. At $t=1$: $$\tan^{-1}(\sqrt2+1)+\tan^{-1}(\sqrt2-1).$$ Using $$(\sqrt2+1)(\sqrt2-1)=1,$$ both positive, so $$\tan^{-1}(\sqrt2+1)+\tan^{-1}(\sqrt2-1)=\frac\pi2.$$ At $t=-1$: $$\tan^{-1}(1-\sqrt2)+\tan^{-1}(-(\sqrt2+1))=-\frac\pi2.$$ Therefore the total bracket value is $$\frac\pi2-\left(-\frac\pi2\right)=\pi.$$ So, $$I=\frac{\sqrt2}{4}\cdot \pi=\frac{\pi}{2\sqrt2}.$$ 8. Hence the value of the integral is $$\boxed{\frac{\pi}{2\sqrt2}}.$$ This corresponds to **Option B**.

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