Question

If $\int \frac{1-{\left(\cot x\right)}^{2010}}{\tan x+{\left(\cot x\right)}^{2011}}dx=\frac{1}{k}{\log }_e|{\left(\sin x\right)}^k+{\left(\cos x\right)}^k|+C$, then $k$ is equal to

• A. $2010$
• B. $2011$
• C. $2012$
• D. $2013$

Answer 1

The correct option is C $2012$

Let $I=\int \frac{1-(\cot x{)}^{2010}}{\tan x+(\cot x{)}^{2010}}dx$

$=\int \frac{{\sin }^{2011}x-{\cos }^{2010}x\sin x}{\tan x{\sin }^{2011}x+{\cos }^{2011}x}$

$=\int \frac{{\sin }^{2011}x\cos x-{\cos }^{2010}x\sin x}{{\sin }^{2012}x+{\cos }^{2012}}$

Let ${\sin }^{2012}x+{\cos }^{2012}x=t$

$dt=2012({\sin }^{2011}x\cos x-{\cos }^{2011}x\sin x)$

$I=\frac{1}{2012}\int \frac{dt}{t}$

$I=\frac{1}{2012}ln|t|+C$

$I=\frac{1}{2012}ln({\sin }^{2012}x+{\cos }^{2012}x)+C$