0 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 1.0k views answer comment Share Follow Print See all 2 Comments 2 2 Comments reply `JEET commented Oct 2, 2019 reply Follow flag $0$? 3 3 replyShare ankitgupta.1729 commented Oct 2, 2019 reply Follow flag yes 2 2 replyShare Please log in or register to add a comment.
3 3 votes It’s a odd function . because sin(-x) = – sinx. anon1 answered Nov 14, 2020 anon1 comment Share Follow 0 reply Please log in or register to add a comment.