Question

If

$\int \frac{\cos^{2}x}{\sin^{6}x}dx = A\cot^{5}x + B\cot^{3}x + k$,

then $A+B$ equals

A $15/8$ B $-15/8$

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Answer


The correct option is B.

To solve this, we start with the given integral:

$$ \int \frac{\cos^2 x}{\sin^6 x} , dx $$

Rewrite the integrand in terms of $\cot x$ and $\csc x$:

$$ \int \cot^2 x \csc^4 x , dx $$

Next, express $\csc^4 x$ in terms of $\cot x$:

$$ \int \cot^2 x (1 + \cot^2 x) \csc^2 x , dx $$

Now, let $t = \cot x$. Then $dt = -\csc^2 x , dx$, which implies:

$$ dx = -\frac{dt}{\csc^2 x} $$

Substitute $t$ into the integral:

$$ I = -\int t^2 (1 + t^2) , dt $$

Expand and integrate:

$$ I = -\int (t^2 + t^4) , dt $$

Integrate each term separately:

$$ I = -\left( \int t^4 , dt + \int t^2 , dt \right) $$

Using standard integration formulas:

$$ I = -\left( \frac{t^5}{5} + \frac{t^3}{3} \right) + k $$

Rewriting it, we get:

$$ \int \frac{\cos^2 x}{\sin^6 x} , dx = -\left( \frac{\cot^5 x}{5} + \frac{\cot^3 x}{3} \right) + k $$

Observing the coefficients, we identify:

$$ A = -\frac{1}{5}, \quad B = -\frac{1}{3} $$

Calculate $A + B$:

$$ A + B = -\frac{1}{5} - \frac{1}{3} $$

To combine these fractions, find a common denominator:

$$ A + B = -\frac{3}{15} - \frac{5}{15} = -\frac{8}{15} $$

Hence,

$$ \boxed{-\frac{8}{15}} $$

Therefore, $A + B = -\frac{8}{15}$. So the correct option is B.


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