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If ∫√1+2tanx(tanx+secx)dx=ln|f(x)|+C, then the value of f(0) is equal to (0≤x<π2)(where C is constant of integration ) |
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Answer» If ∫√1+2tanx(tanx+secx)dx=ln|f(x)|+C, then the value of f(0) is equal to (0≤x<π2) (where C is constant of integration ) |
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