0 votes 0 votes The value of the integral $\displaystyle{}\int_{-1}^1 \dfrac{x^2}{1+x^2} \sin x \sin 3x \sin 5x dx$ is $0$ $\frac{1}{2}$ $ – \frac{1}{2}$ $1$ Calculus isi2014-dcg calculus integration definite-integral + – Arjun asked Sep 23, 2019 recategorized Nov 12, 2019 by Lakshman Bhaiya Arjun 587 views answer comment Share Follow See all 2 Comments See all 2 2 Comments reply `JEET commented Oct 2, 2019 reply Follow Share $0$? 3 votes 3 votes ankitgupta.1729 commented Oct 2, 2019 reply Follow Share yes 2 votes 2 votes Please log in or register to add a comment.
2 votes 2 votes It’s a odd function . because sin(-x) = – sinx. raja11sep answered Nov 14, 2020 raja11sep comment Share Follow See all 0 reply Please log in or register to add a comment.