Rinkoo Gupta
Last Activity: 10 Years ago
Integral (lim o to 2 pie) e^x.sin([ie/4+x/2) dx
let pie/4+x/2=t
on diff w r to x we get dx = 2dt
new limits are pie/4 and 5pie/4
integral(lim pie/4 to 5pie/4)e^2(t-pie/4) .sint 2 dt
= 2 e^-pie/2 integral (pie/4 to 5pie/4) e^2t sint dt
=2e^-pie/2[e^2t(2sint+cost]/4+1
=2/5. e^-pie/2[e^5pie/3{2 sin5pie/4-cos5pie/4}-e^2pie/4{2sinpie/4-cospie/4}]
=2/5e^-pie/2[cospie/4(e^5pie/2+e^pie/2)-2 sin(e^5pie/2+e^pie/2)]
Thanks & Regards
Rinkoo Gupta
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