Question

The integral $\int \frac{1-(\cot x{)}^{2008}}{\tan x+(\cot x{)}^{2009}}dx$ is equal to
($C$ is constant of integration)

• A. $\frac{1}{2010}\ln \left|\right(\sin x{)}^{2011}+\left(\cos x{)}^{2011}\right|+C$
• B. $\frac{1}{2011}\ln \left|\right(\sin x{)}^{2010}+\left(\cos x{)}^{2010}\right|+C$
• C. $\frac{1}{2010}\ln \left|\right(\sin x{)}^{2010}+\left(\cos x{)}^{2010}\right|+C$
• D. $\frac{1}{2011}\ln \left|\right(\sin x{)}^{2011}+\left(\cos x{)}^{2011}\right|+C$

Answer 1

The correct option is C $\frac{1}{2010}\ln \left|\right(\sin x{)}^{2010}+\left(\cos x{)}^{2010}\right|+C$

Let $I=\int \frac{1-(\cot x{)}^{2008}}{\tan x+(\cot x{)}^{2009}}dx$

$\Rightarrow I=\int \frac{{\sin }^{2009}x\cos x\left[\right(\sin x{)}^{2008}-\left(\cos x{)}^{2008}\right]}{{\sin }^{2008}x\left[\right(\sin x{)}^{2010}+\left(\cos x{)}^{2010}\right]}dx\Rightarrow I=\int \frac{(\sin x{)}^{2009}\cos x-(\cos x{)}^{2009}\sin x}{(\sin x{)}^{2010}+(\cos x{)}^{2010}}dx$

Assuming $(\sin x{)}^{2010}+(\cos x{)}^{2010}=t$
$\Rightarrow 2010\left[\right(\sin x{)}^{2009}\cos x-\left(\cos x{)}^{2009}\sin x\right]dx=dt$
Therefore,
$I=\frac{1}{2010}\int \frac{dt}{t}=\frac{1}{2010}\ln \left|\right(\sin x{)}^{2010}+\left(\cos x{)}^{2010}\right|+C$