Get Study Tools
Answer & Explanation
2022-01-04Added 42 answers
2022-01-05Added 34 answers
2022-01-11Added 439 answers
∫ecosxdx=∫n=0∞cos2nx(2n)!dx+∫∑n=0∞cos2n+1x(2n+1)!dx=∫(1+∑n=1∞cos2nx(2n)!)dx+∫∑n=0∫cos2n+1x(2n+1)!dxFor n is any natural number,∫cos2nxdx=(2n)!x4n(n!)2+∑k=1n(2n)!((k−1)!)2sinxcos2k−1x4n−k+1(n!)2(2k−1)!+CThis result can be done by successive integration by parts.For n is any non-negative integer,∫cos2n+1xdx=∫cos2nxd(sinx)=∫(1−sin2x)nd(sinx)=∫∑k=0∞Ckn(−1)ksin2kxd(sinx)=∑k=0n(−1)kn!sin2k+1xk!(n−k)!(2k+1)+C∴∫(1+∑n=1∞cos2nx(2n)!)dx+∫∑n=0∞cos2n+1x(2n+1)!dx=x+∑n=1∞x4n(n!)2+∑n=1∞∑k=1n(k−1)!)2sinxcos2k−1x4n−k+1(n!)2(2k−1)!+∑k=0n(−1)kn!sin2k+1x(2n+1)!k!(n−k)!(2k+1)+C
Do you have a similar question?
Recalculate according to your conditions!
Ask your question. Get your answer.
Let our experts help you. Answer in as fast as 15 minutes.
Most Popular Questions
Didn't find what you were looking for?
Connect with us
Get Plainmath App
Louki Akrita, 23, Bellapais Court, Flat/Office 46, 1100, Nicosia, Cyprus
Cyprus reg.number: ΗΕ 419361
E-mail us: firstname.lastname@example.org
Our Service is useful for:
Plainmath is a platform aimed to help users to understand how to solve math problems by providing accumulated knowledge on different topics and accessible examples. Plainmath.net is owned and operated by RADIOPLUS EXPERTS LTD.
2022 Plainmath. All rights reserved
Get answers within minutes and finish your homework faster
Dont have an account?
Create a free account to see answers
Already have an account?