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 30 National Universities for Engineering (NIRF 2024) | 30+ Specializations | AI Powered Learning & State-of-the-Art Facilities
Amrita University B.Tech 2026
Apply
Recognized as Institute of Eminence by Govt. of India | NAAC ‘A++’ Grade | Upto 75% Scholarships
Amity University, Noida | Law...
Apply
700+ Campus placements at top national and global law firms, corporates and judiciaries
Great Lakes Institute of Mana...
Apply
Admissions Open | Globally Recognized by AACSB (US) & AMBA (UK) | 17.8 LPA Avg. CTC for PGPM 2025
Manav Rachna University Law A...
Apply
Admissions open for B.A. LL.B. (Hons.), B.B.A. LL.B. (Hons.) and LL.B Program (3 Years) | School of Law, MRU ranked No. 1 in Law Schools of Excelle...
Nirma University Law Admissio...
Apply
Grade 'A+' accredited by NAAC | Ranked 33rd by NIRF 2025
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