What is the integral from 0 to tan(x) of 1/(1+t^2) ?

The integral from 0 to tan(x) of 1/(1+t^2) is arctan(tan(x))

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integral from 0 to tan(x) of 1/(1+t^2)

\int_{0}^{t a n (x)} \frac{1}{1 + t ^{2}} d t

arctan (tan (x))

Solution steps

\int_{0}^{t a n (x)} \frac{1}{1 + t ^{2}} d t

Use the common integral:

\int \frac{1}{1 + t ^{2}} d t = arctan (t)

= [arctan (t)]_{0}^{t a n (x)}

Compute the boundaries:

arctan (tan (x))

= arctan (tan (x))

\frac{x ^{2} - x - 2}{x ^{2} - 6 x + 9}

\frac{d}{d x} (arctan (5^{x}))

\int \frac{1 - x}{x} d x

x \to 6 + lim (ln (x - 6))

y = sin (x) + x cos (x)

What is the integral from 0 to tan(x) of 1/(1+t^2) ?
The integral from 0 to tan(x) of 1/(1+t^2) is arctan(tan(x))