Question

What is the value of \(\rm \int_0^{\pi/4}(\cot^3 x + \cot x)dx \ ?\)
1. \(-\dfrac{1}{4}\)
2. \(\dfrac{1}{2}\)
3. -\(\dfrac{1}{2}\)
4. None of these

Answer 1

Correct Answer - Option 1 : \(-\dfrac{1}{4}\)

Concept:

cot² x = cosec² x - 1 

\(\rm \int x^{n}dx = \frac{x^{n+1}}{n+1} + C\) 

Calculation:

I = \(\rm \int_0^{π/4}(\cot^3 x + \cot x)dx \ \) 

I = \(\rm\int_{0}^{π /4}\left [ \left ( cosec^{2}x-1 \right )cotx+ cotx \right ]dx\) 

I = \(\rm \int_{0}^{π /4}\left [ cosec^{2}x. cotx-cotx+cotx \right ]dx\) 

I = \(\rm \int_{0}^{π /4}\left ( cosec^{2}x.cotx \right )dx\) 

I = \(\rm \int_{0}^{π /4}\left ( \frac{1}{sin^{2}x} . \frac{cosx}{sinx}\right )dx\) 

I = \(\rm\int_{0}^{π/4}\left ( \frac{cosx}{sin^{3}x} \right )dx\)           ...(1)

Let , sin x = t 

Differentiate both side  w.r.t  x , 

cos x dx = dt   

If , x = 0 , then t = 0 

x = π /4 , then t = \(\frac{1}{\sqrt{2}}\) 

substitute above values in eq. (1) 

I = \(\rm\int_{0}^{\frac{1}{\sqrt{2}}}\left ( \frac{dt}{t^{3}} \right )\)  

I = \(\rm \left [ \frac{1}{-2t^{2}} \right ]_{0}^{\frac{1}{\sqrt{2}}}\) 

I = \(\rm \frac{-1}{2}\left [ \frac{1}{2}-0 \right ]\)  

I = \(\frac{-1}{4}\) 

The correct option is 1.