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Evaluate : `intsin4x.e^(tan^(2)x)dx` |
Answer» Correct Answer - `-2e^(tan^(2)x)cos^(4)x+C` `intsin4x.e^(tan^(2)x)dx=4intsinx cosx 2x e^(tan^(2)x)dx` `=4inttanx.sec^(2)x.cos^(4)x.cos2xe^(tan^(2)x)dx` `2int(1)/((1+t)^(2))(1-t)/(1+t)e^(t)dt` (Putting, `tan^(2)x=t rArr 2 tan x. sec^(2)xdx=dt`) `=-2int((t+1)-2)/((1+t)^(3))e^(t)dt` `=-2inte^(t)((1)/((1+t)^(2))+(-2)/((1+t)^(3)))dt` `=(-2e^(t))/((1+t)^(2))+C` `=-2e^(tan^(2)x)cos^(4)+C` |
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