Indefinite Integration 2 Question 4

4.. Show that 0nπ+v|sinx|dx=2n+1cosv, where n is a positive integer and 0v<π.

(1994, 4M)

Show Answer

Answer:

Correct Answer: 4. (4)

Solution:

  1. 0nπ+v|sinx|dx=0π|sinx|dx+π2π|sinx|dx+

+(n1)πnπ|sinx|dx+nπnπ+v|sinx|dx

=r=1n(r1)πrπ|sinx|dx+nπnπ+v|sinx|dx

Now to solve, (r1)πrπ|sinx|dx, we have

x=(r1)π+tsinx=sin[(r1)π+t]=(1)r1sint

and when x=(r1)π,t=0 and when

x=rπ,t=π

(r1)πrπ|sinx|dx=0π|(1)r1sint|dt

=0π|sint|dt=0πsintdt=[cost]0π=cosπ+cos0=2

Again, nπnπ+v|sinx|dx, putting x=nπ+t

Then, nπnπ+v|sinx|dx=0v|(1)nsint|dt=0vsintdt

=[cost]0v=cosv+cos0=1cosv

0nπ+v|sinx|dx=r=1n(r1)πrπlnsinx|dx+nπnπ+v|sinxdx

=r=1n2+nπnπ+v|sinx|dx=2n+1cosv



NCERT Chapter Video Solution

Dual Pane