The value of ∫ cosec6 x dx will be ___________, where c is an arbit

1.	The value of ∫ cosec6 x dx will be ___________, where c is an arbitrary constant.1. \(\frac {-1}5 \cot^5x-\frac 2 3 \cot^3 x - \cot x + c\)2. -sin7 x - 5 sin5 x - 3 sin3 x - sin x + c3. cos5 x + 3 cos3 x + cos x + c4. cos6 x - 4 cos4 x + 2 cos2 x + c
Answer» Correct Answer - Option 1 : \(\frac {-1}5 \cot^5x-\frac 2 3 \cot^3 x - \cot x + c\) I = ∫ cosec6 x dx I = ∫ cosec⁴ x cosec² x dx We know that: cosec² x = 1 + cot²x I = ∫ cosec⁴x (1 + cot² x) dx I = ∫ cosec⁴x dx + ∫ cot² x cosec⁴x dx I = ∫ cosec²x cosec²x dx + ∫ cot² x (cosec²x cosec² x) dx I = ∫ cosec²x (1 + cot² x) dx + ∫ cot² x cosec²x (1 + cot² x) dx I = ∫ cosec²x dx + ∫ cosec² x cot² x dx + ∫ cot² x cosec² x + ∫ cot⁴x cosec² x dx I = ∫ cosec2 x dx + 2 ∫ cosec2 x cot2 x dx + ∫ cot4 x cosec2 x dx Put cot x = z By differentiating we have, - cosec² x dx = dz I = ∫ (- dz) + 2 ∫ z² (- dz) + ∫ z⁴(- dz) \(I~=~-~z~-~\frac{2z^3}{3}~-~\frac{z^5}{5}~+~c\) Put z = cot x in above equation: \(I~=~-~cot~x~-~\frac{2cot^3~x}{3}~-~\frac{cot^5~x}{5}~+~c\) \(I~=~\frac {-1}5 \cot^5x-\frac 2 3 \cot^3 x - \cot x + c\)

The value of ∫ cosec6 x dx will be ___________, where c is an arbitrary constant.1. \(\frac {-1}5 \cot^5x-\frac 2 3 \cot^3 x - \cot x + c\)2. -sin7 x - 5 sin5 x - 3 sin3 x - sin x + c3. cos5 x + 3 cos3 x + cos x + c4. cos6 x - 4 cos4 x + 2 cos2 x + c

Answer» Correct Answer - Option 1 : \(\frac {-1}5 \cot^5x-\frac 2 3 \cot^3 x - \cot x + c\)

I = ∫ cosec6 x dx

I = ∫ cosec⁴ x cosec² x dx

We know that:

cosec² x = 1 + cot²x

I = ∫ cosec⁴x (1 + cot² x) dx

I = ∫ cosec⁴x dx + ∫ cot² x cosec⁴x dx

I = ∫ cosec²x cosec²x dx + ∫ cot² x (cosec²x cosec² x) dx

I = ∫ cosec²x (1 + cot² x) dx + ∫ cot² x cosec²x (1 + cot² x) dx

I = ∫ cosec²x dx + ∫ cosec² x cot² x dx + ∫ cot² x cosec² x + ∫ cot⁴x cosec² x dx

I = ∫ cosec2 x dx + 2 ∫ cosec2 x cot2 x dx + ∫ cot4 x cosec2 x dx

Put cot x = z

By differentiating we have,

- cosec² x dx = dz

I = ∫ (- dz) + 2 ∫ z² (- dz) + ∫ z⁴(- dz)

\(I~=~-~z~-~\frac{2z^3}{3}~-~\frac{z^5}{5}~+~c\)

Put z = cot x in above equation:

\(I~=~-~cot~x~-~\frac{2cot^3~x}{3}~-~\frac{cot^5~x}{5}~+~c\)

\(I~=~\frac {-1}5 \cot^5x-\frac 2 3 \cot^3 x - \cot x + c\)