The value of the integral _-( )/(4)^( )/(4)(32 ^4x1+e^ x)dx is :

Question

The value of the integral $\int _{-\frac{\pi }{4}}^{\frac{\pi }{4}}(\frac{32{\cos }^{4}x}{1+{e}^{\sin x}})dx$ is :

Accepted Answer

Let

$I={\int }_{\frac{\pi }{4}}^{-\frac{\pi }{4}}\frac{32{\cos }^{4}x}{1+{e}^{\sin x}}\, dx$

We will use the standard property:

If

$I={\int }_{a}^{-a}f(x)\, dx,$

then

$I={\int }_{a}^{-a}f(-x)\, dx$

and hence

$2I={\int }_{a}^{-a}(f(x)+f(-x))\, dx$

Now here,

$f(x)=\frac{32{\cos }^{4}x}{1+{e}^{\sin x}}$

So,

$f(-x)=\frac{32{\cos }^{4}(-x)}{1+{e}^{\sin (-x)}}$

Using $\cos (-x)=\cos x$ and $\sin (-x)=-\sin x$,

$f(-x)=\frac{32{\cos }^{4}x}{1+{e}^{-\sin x}}$

Therefore,

$f(x)+f(-x)=32{\cos }^{4}x(\frac{1}{1+{e}^{\sin x}}+\frac{1}{1+{e}^{-\sin x}})$

Now simplify the bracket:

$\frac{1}{1+{e}^{\sin x}}+\frac{1}{1+{e}^{-\sin x}}$

Let $t={e}^{\sin x}$. Then ${e}^{-\sin x}=\frac{1}{t}$. So,

$\frac{1}{1+t}+\frac{1}{1+\frac{1}{t}}=\frac{1}{1+t}+\frac{t}{1+t}=1$

Hence,

$f(x)+f(-x)=32{\cos }^{4}x$

So,

$2I={\int }_{\frac{\pi }{4}}^{-\frac{\pi }{4}}32{\cos }^{4}x\, dx$

Thus,

$I=16{\int }_{\frac{\pi }{4}}^{-\frac{\pi }{4}}{\cos }^{4}x\, dx$

Since ${\cos }^{4}x$ is an even function,

$I=32{\int }_{\frac{\pi }{4}}^{0}{\cos }^{4}x\, dx$

Now use the identity

${\cos }^{4}x={({\cos }^{2}x)}^{2}={(\frac{1+\cos 2x}{2})}^{2}$

which gives

${\cos }^{4}x=\frac{3+4\cos 2x+\cos 4x}{8}$

Therefore,

$I=32{\int }_{\frac{\pi }{4}}^{0}\frac{3+4\cos 2x+\cos 4x}{8}\, dx$

$I=4{\int }_{\frac{\pi }{4}}^{0}(3+4\cos 2x+\cos 4x)\, dx$

Now integrate termwise:

$\int 3\, dx=3x,\, \int 4\cos 2x\, dx=2\sin 2x,\, \int \cos 4x\, dx=\frac{\sin 4x}{4}$

So,

$I=4{[3x+2\sin 2x+\frac{\sin 4x}{4}]}_{\frac{\pi }{4}}^{0}$

At $x=\frac{\pi }{4}$,

$3x=\frac{3\pi }{4},\, \sin 2x=\sin \frac{\pi }{2}=1,\, \sin 4x=\sin \pi =0$

At $x=0$,

$3x=0,\, \sin 0=0$

Hence,

$I=4(\frac{3\pi }{4}+2)$

$I=3\pi +8$

So the correct answer is:

$3\pi +8$

Hence, Option B is correct.

Solutions · 1