Question

If $\int \frac{{\cos }^{4}x}{{\sin }^{2}x}dx=A\cot x+B\sin 2x+C\frac{x}{2}+D$, then

• A. $A=-2,B=1/4$
• B. $B=-1/4,C=-3$
• C. $B=1/4,C=-3$
• D. None of these

Answer 1

Answer: (B)

$B=-1/4,C=-3$

Now,

$\int \frac{{\cos }^{4}x}{{\sin }^{2}x}dx$

$=\int \frac{({\cos }^{2}x{)}^2}{{\sin }^{2}x}dx$

$=\int \frac{(1-{\sin }^{2}x{)}^2}{{\sin }^{2}x}dx$

$=\int \frac{(1-2{\sin }^{2}x+{\sin }^{4}x)}{{\sin }^{2}x}dx$

$=\int (cose{c}^{2}x-2+{\sin }^{2}x)dx$

$=\int (cose{c}^{2}x-2+\frac{1-\cos 2x}{2})dx$

$=-\cot x-2x+\frac{x}{2}-\frac{\sin 2x}{4}+c$ [ Where $c$ is integrating constant]

$=-\cot x-\frac{3x}{2}-\frac{\sin 2x}{4}+c$

Comparing this with the given expression we get,

$A=-1,B=-\frac{1}{4},C=-3.$ , D = c