1. Tardigrade
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  4. If the integral ∫ ( cos 8 x+1/ cot 2 x- tan 2 x) d x=A cos 8 x+k, where k is an arbitrary constant, then A is equal to :

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Q. If the integral $\int \frac{\cos 8 x+1}{\cot 2 x-\tan 2 x} d x=A \cos 8 x+k$, where $k$ is an arbitrary constant, then $A$ is equal to :

Integrals

Solution:

Let $I =\int \frac{\cos 8 x+1}{\cot 2 x-\tan 2 x} d x$
Now, $D ^{r}=\cot 2 x-\tan 2 x=\frac{\cos 2 x}{\sin 2 x}-\frac{\sin 2 x}{\cos 2 x}$
$=\frac{\cos ^{2} 2 x-\sin ^{2} 2 x}{\sin 2 x \cos 2 x}=\frac{2 \cos 4 x}{\sin 4 x}$
$\therefore I =\int \frac{2 \cos ^{2} 4 x}{\frac{2 \cos 4 x}{\sin 4 x}} d x$
$=\int \frac{2 \cos ^{2} 4 x \cdot \sin 4 x}{2 \cos 4 x} d x$
$=\frac{1}{2} \int \sin 8 x d x=-\frac{1}{2} \frac{\cos 8 x}{8}+k$
$=-\frac{1}{16} \cdot \cos 8 x +k$
Now, $-\frac{1}{16} \cdot \cos 8 x +k= A \cos 8 x +k$
$\Rightarrow A =-\frac{1}{16}$