Question

If $I_n=\int {\cot }^{n}xdx,$ then $I_0+I_1+2(I_2+I_3)+I_4+I_5=$
(where $C$ is integration constant)

• A. $-[\cot x+\frac{{\cot }^{2}x}{2}+\frac{{\cot }^{3}x}{3}+\frac{{\cot }^{4}x}{4}]+C$
• B. $-[\cot x+\frac{{\cot }^{3}x}{3}]+C$
• C. $-[\frac{{\cot }^{4}x}{4}+\frac{{\cot }^{2}x}{2}]+C$
• D. $-[\cot x+\frac{{\cot }^{3}x}{3}+\frac{{\cot }^{5}x}{5}]+C$

Answer 1

The correct option is A $-[\cot x+\frac{{\cot }^{2}x}{2}+\frac{{\cot }^{3}x}{3}+\frac{{\cot }^{4}x}{4}]+C$

For, $\int {\cot }^{n}xdx,n\ge 2$
$I_n+I_{n-2}=-\frac{{\cot }^{n-1}x}{n-1}+C$
Now, on putting $n=2,3,4$ and $5$ in the above expression;
we have,
$I_2+I_0=-\cot x+C_0{I}_3+I_1=-\frac{{\cot }^{2}x}{2}+C_1{I}_4+I_2=-\frac{{\cot }^{3}x}{3}+C_2{I}_5+I_3=-\frac{{\cot }^{4}x}{4}+C_3$
Now adding all, we get
$I_0+I_1+2(I_2+I_3)+I_4+I_5=-[\cot x+\frac{{\cot }^{2}x}{2}+\frac{{\cot }^{3}x}{3}+\frac{{\cot }^{4}x}{4}]+C$