Question:
If the integral
$\int \frac{cos 8x+1}{cot 2x-tan 2x} dx=A cos 8x+k,$
where $k$ is an arbitrary constant, then $A$ is equal to:
If the integral
$\int \frac{cos 8x+1}{cot 2x-tan 2x} dx=A cos 8x+k,$
where $k$ is an arbitrary constant, then $A$ is equal to:
Updated On: Nov 21, 2025
- $-\frac{1}{16}$
- $\frac{1}{16}$
- $\frac{1}{8}$
- $-\frac{1}{8}$
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The Correct Option is A
Solution and Explanation
$Let I=\int \frac{cos \, 8x+1}{cot \, 2x-tan\, 2x} dx$
$Now, D^{r}=cot 2x -tan 2x =\frac{cos 2x}{sin 2x}-\frac{sin 2x}{cos 2x}$
$=\frac{cos^{2}2x-sin^{2} 2x}{sin 2x cos 2x}=\frac{2 cos 4x}{sin 4x}$
$\therefore I=\int \frac{2 cos ^{2} 4x}{\frac{2 cos 4x}{sin 4x}} dx=\int \frac{2 cos ^{2}4x. sin 4x}{2 cos 4x} dx$
$=\frac{1}{2} \int sin 8x dx=-\frac{1}{2} \frac{cos 8x}{8}+k$
$=-\frac{1}{16}. cos 8x +k$
$Now, -\frac{1}{16}. cos 8x+k=A cos 8x+k$
$\Rightarrow A=-\frac{1}{16}$
$Now, D^{r}=cot 2x -tan 2x =\frac{cos 2x}{sin 2x}-\frac{sin 2x}{cos 2x}$
$=\frac{cos^{2}2x-sin^{2} 2x}{sin 2x cos 2x}=\frac{2 cos 4x}{sin 4x}$
$\therefore I=\int \frac{2 cos ^{2} 4x}{\frac{2 cos 4x}{sin 4x}} dx=\int \frac{2 cos ^{2}4x. sin 4x}{2 cos 4x} dx$
$=\frac{1}{2} \int sin 8x dx=-\frac{1}{2} \frac{cos 8x}{8}+k$
$=-\frac{1}{16}. cos 8x +k$
$Now, -\frac{1}{16}. cos 8x+k=A cos 8x+k$
$\Rightarrow A=-\frac{1}{16}$
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Concepts Used:
Integrals of Some Particular Functions
There are many important integration formulas which are applied to integrate many other standard integrals. In this article, we will take a look at the integrals of these particular functions and see how they are used in several other standard integrals.
Integrals of Some Particular Functions:
- ∫1/(x2 – a2) dx = (1/2a) log|(x – a)/(x + a)| + C
- ∫1/(a2 – x2) dx = (1/2a) log|(a + x)/(a – x)| + C
- ∫1/(x2 + a2) dx = (1/a) tan-1(x/a) + C
- ∫1/√(x2 – a2) dx = log|x + √(x2 – a2)| + C
- ∫1/√(a2 – x2) dx = sin-1(x/a) + C
- ∫1/√(x2 + a2) dx = log|x + √(x2 + a2)| + C
These are tabulated below along with the meaning of each part.
