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Question Number 83781 by john santu last updated on 06/Mar/20
∫0π214sin2x+5cos2xdx
Commented by niroj last updated on 06/Mar/20
∫0π214sin2x+5cos2xdx=∫0π2sec2xdx4tan2x+5puttanx=tsec2xdx=dtifx=π2thent=∞ifx=0thent=0=∫0∞14t2+5dt=14∫0∞1t2+54dt=14[∫1(t)2+(52)2dt]0∞=14[152tan−1t52]0∞=14[25tan−1(∞).25−0]=14.25.π2=π45//.
Commented by john santu last updated on 06/Mar/20
inshortcut=∫0π2dxa2sin2x+b2cos2x=π2ab[a=4=2,b=5]=π2.2.5=π45★
Answered by MJS last updated on 06/Mar/20
∫dx4sin2x+5cos2x=[t=tanx→dx=dtt2+1;sinx=tt2+1;cosx=1t2+1]=∫dt4t2+5=510arctan25t5==510arctan25tanx5+Csnswerisπ520
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