calculate-0-pi-1-2sinx-3-2cosx-dx-let-A-0-pi-1-2sinx-3-2cosx-dx-changement-tan-x-2-t-give-A-0-1-4t-1-t-2-3-2-1-t-2-1-t-2-2dt-1-t-2-2-0- Tinku Tara June 4, 2023 Integration 0 Comments FacebookTweetPin Question Number 53112 by maxmathsup by imad last updated on 21/Jan/19 calculate∫0π1+2sinx3+2cosxdxletA=∫0π1+2sinx3+2cosxdxchangementtan(x2)=tgiveA=∫0∞1+4t1+t23+21−t21+t22dt1+t2=2∫0∞1+t2+4t(1+t2)2(3+3t2+2−2t21+t2)dt=2∫0∞t2+4t+1(1+t2)(5+t2)dtletdecomposeF(t)=t2+4t+1(t2+1)(t2+5)F(t)=at+bt2+1+ct+dt2+5⇒(at+b)(t2+5)+(ct+d)(t2+1)=t2+4t+1⇒at3+5at+bt2+5b+ct3+ct+dt2+d=t2+4t+1⇒(a+c)t3+(b+d)t2+(5a+c)t+5b+d=t2+4t+1⇒a+c=0andb+d=1and5a+c=4and5b+d=1⇒c=−a⇒a=1⇒c=−1wehaved=1−b⇒5b+1−b=1⇒b=0⇒d=1⇒F(t)=tt2+1+−t+1t2+5⇒A=2∫0∞F(t)dt=∫0∞2tt2+1dt+∫0∞−2t+2t2+5dt=[ln(t2+1t2+5)]0+∞+2∫0∞dtt2+5=ln(5)+2∫0∞dtt2+5but∫0∞dtt2+5dt=t=5u∫0∞5du5(1+u2)=15[artanu]0+∞=π25⇒A=ln(5)+π25. Answered by tanmay.chaudhury50@gmail.com last updated on 19/Jan/19 ∫13+2cosxdx+∫2sinx3+2cosxdx∫1+tan2x23+3tan2x2+2−2tan2x2dx−∫d(3+2cosx)3+2cosx∫sec2x25+tan2x2dx−ln(3+2cosx)+c2∫d(tanx2)5+tan2x2−do∣2×15tan−1(tanx25)−ln(3+2cosx)∣0π[{25tan−1(tanπ25)−ln(3+2cosπ)}−{25tan−1(tan05)−ln(3+2cos0)}]=[{25(π2)−ln(3−2)}−{25tan−1(0)−ln5}]=π5+ln5 Terms of Service Privacy Policy Contact: info@tinkutara.com FacebookTweetPin Post navigation Previous Previous post: x-2-1-3-x-3-1-3-gt-1-2-Next Next post: let-I-t-1-t-2-t-1-2-dt-find-value-of-I- Leave a Reply Cancel replyYour email address will not be published. Required fields are marked *Comment * Name * Save my name, email, and website in this browser for the next time I comment.