Evaluate the following integrals:

Solution:

Let $I=\int \frac{\cos ^{2} x-\sin ^{2} x}{\sqrt{1+\cos 4 x}} d x$

We know $\cos 2 \theta=2 \cos ^{2} \theta-1=\cos ^{2} \theta-\sin ^{2} \theta$

Hence, in the numerator, we can write $\cos ^{2} x-\sin ^{2} x=\cos 2 x$

In the denominator, we can write $4 x=2 \times 2 x$

$\Rightarrow 1+\cos 4 x=1+\cos (2 \times 2 x)$

$\Rightarrow 1+\cos 4 x=2 \cos ^{2} 2 x$

Therefore, we can write the integral as

$\mathrm{I}=\int \frac{\cos 2 \mathrm{x}}{\sqrt{2 \cos ^{2} 2 \mathrm{x}} \mathrm{dx}}$

$\Rightarrow \mathrm{I}=\int \frac{\cos 2 \mathrm{x}}{\sqrt{2} \cos 2 \mathrm{x}} \mathrm{dx}$

$\Rightarrow \mathrm{I}=\int \frac{1}{\sqrt{2}} \mathrm{dx}$

$\Rightarrow \mathrm{I}=\frac{1}{\sqrt{2}} \int \mathrm{dx}$

Recall $\int \mathbf{d x}=\mathbf{x}+\mathbf{c}$

$\Rightarrow I=\frac{1}{\sqrt{2}} \times x+c$

$\therefore I=\frac{x}{\sqrt{2}}+c$

Thus, $\int \frac{\cos ^{2} x-\sin ^{2} x}{\sqrt{1+\cos 4 x}} d x=\frac{x}{\sqrt{2}}+c$