68 Views

evaluate integration of x^2sinx with respect to x


MASTER TECHY IN KANNADA 13th Apr, 2021
Answer (1)
Ujjwal Kumawat 13th Apr, 2021

Dear Aspirant

I=∫x^2sinx.dx

Applying integration by part,
I=x^2∫sinx.dx−∫[dxd(x^2)∫sindx]dx
=x^2(−cosx)−2∫x(−cosx).dx
=−x^2cosx+2∫xcosxdx
Again applying integration by parts
I=−x^2cosx+2[x∫cosx.dx−∫dxd(x)∫cosx.dx]
=−x^2cosx+2[xsinx−∫sinx.dx ]
I=−x2cosx+2xsinx+2cosx+c

i hope it will help you

Thank you

Related Questions

Amity University-Noida B.Tech...
Apply
Among top 100 Universities Globally in the Times Higher Education (THE) Interdisciplinary Science Rankings 2026
Amity University-Noida M.Tech...
Apply
Among top 100 Universities Globally in the Times Higher Education (THE) Interdisciplinary Science Rankings 2026
Amity University-Noida MBA Ad...
Apply
Ranked among top 10 B-Schools in India by multiple publications | Top Recruiters-Google, MicKinsey, Amazon, BCG & many more.
Amity University-Noida BBA Ad...
Apply
Among top 100 Universities Globally in the Times Higher Education (THE) Interdisciplinary Science Rankings 2026
RV University, Mysuru | B.Tec...
Apply
World-class and highly qualified engineering faculty. High-quality global education at an affordable cost
New Horizon College BBA Admis...
Apply
UG Admissions 2026 open| NAAC ‘A’ grade | Merit-based Scholarships available.
View All Application Forms

Download the Careers360 App on your Android phone

Regular exam updates, QnA, Predictors, College Applications & E-books now on your Mobile

150M+ Students
30,000+ Colleges
500+ Exams
1500+ E-books