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VIKRAM KUMAR Grade: 12
        Integral  1+tan(x)tan(x+a)
6 years ago

Answers : (1)

vikas askiitian expert
510 Points
										

I = [1 + tanxtanx+a]dx


I = [1 + (sinx/cosx)(sinx+a/cosx+a)]


I = [cosx+acosx+sinx+asinx]dx/cosxcosx+a                  (using , cosacosb+sinasinb = cosa-b)


I = (cosa)dx/cosxcosx+a 


multiply divide by sina


I = cota (sina)dx/cosxcosx+a          ...............1


sina = sin[(x+a)-x] = sin(x+a)cosx - cos(x+a)sinx        ............2


putting 2 in eq 1


I = cota [ tan(x+a) - tanx]dx


I = cota [ logsec(x+a) - logsecx] + c


I = cota { log[sec(x+a)/secx] } + c


this is the required integral


approve my ans if u like it

6 years ago
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