∫√(1+x)/(1-x)dx ∫sinxcosx/4+(cosx)^4
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∫√(1+x)/(1-x)dx ∫sinxcosx/4+(cosx)^4
∫√[(1+x)/(1-x)] dx=∫(1+x)/√(1-x^2) dx
=arcsinx-∫1/2√(1-x^2) d(1-x^2)
=arcsinx-√(1-x^2)+C
∫sinxcosx/[4+(cosx)^4] dx=∫-cosx/[4+(cosx)^4] dcosx
=∫-t/(4+t^4) dt
=∫-1/2(4+t^4) dt^2
=-1/4∫1/[1+(t^2/2)^2] dt^2/2
=-arctan(t^2/2)/4+C
=-arctan(cos^2 x /2)/4+C
=arcsinx-∫1/2√(1-x^2) d(1-x^2)
=arcsinx-√(1-x^2)+C
∫sinxcosx/[4+(cosx)^4] dx=∫-cosx/[4+(cosx)^4] dcosx
=∫-t/(4+t^4) dt
=∫-1/2(4+t^4) dt^2
=-1/4∫1/[1+(t^2/2)^2] dt^2/2
=-arctan(t^2/2)/4+C
=-arctan(cos^2 x /2)/4+C
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