Question

integration of. tan.inverse.x.dx​

Answer 1

Step-by-step explanation:

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∫ tan^(-1)x dx

Let tan^(-1)x = y , then tany = x .....(1)

Now, differentiating both sides w.r.t "y" :

sec²y * (dy/dx) = 1

dy/dx = 1/sec²y

dy*sec²y = dx

So, ∫ tan^(-1)x dx = ∫y*sec²y dy

Let f(y) = y

f'(y) = 1

Let g'(y) = sec²y

g(y) = tany

We know that :

∫f(y)*g'(y)dy = f(y)d(y) - ∫f'(y)*g(y)dy

So,

We know that :

∫f'(x) / f(x)dx = ln |f(x)| + c

∫y*sec²y dy

= ysec²y - ∫tanydy

= ysec²y - ∫siny/(cosy)dy

= ysec²y + ∫-siny/(cosy)dy

= [tan(-1)x][1+x²] + ln |cosy| + c

= [tan(-1)x][1+x²] -(1/2) ln |(x²+1)| + c

Because :

tan²y = x²

sec²y -1 = x²

sec²y = x² +1

secy = √(x² + 1)

cosy = 1/√(x² + 1)