Use app×
Join Bloom Tuition
One on One Online Tuition
JEE MAIN 2025 Foundation Course
NEET 2025 Foundation Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
0 votes
128 views
in Sets, Relations and Functions by (94.9k points)
closed by
`int_0^pi(xtanx)/(tanx+secx).dx=[pi(pi-2)]/2`

1 Answer

0 votes
by (95.5k points)
selected by
 
Best answer
`I=int_0^pi (xtanx)/(tanx+secx)dx`
`=int_0^pi (xsinx/cosx)/(sinx/cosx+1/cosx)dx`
`I=int_0^pi (xsinx)/(1+sinx)-(1)`
`I=int_0^pi((pi-x)sin(pi-x))/(1+sin(pi-x)`
`I=int_0^pi((pi-x)sinx)/(1+sinx)-(2)`
`I=int_0^pi[(xsinx)/(1+sinx)+((pi-x)sinx)/(1+sinx)]dx`
`2I=int_0^pi[(xsinx+pisinx-xsinx)/(1+sinx)]dx`
`2I=pi int_0^pi sinx/(1+sinx)dx`
`I=pi/2 int_0^pi sinx/(1+sinx)dx`
`[secpi-sec0]-int_0^pi sec^2x dx+int_0^pi 1dx`
`-1-1-[tanx]_0^pi+(x)_0^pi`
`-2+pi`
`pi-2`
`I=pi/2(pi-2)`.

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

Categories

...