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

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

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

\int_{2}^{s i n (x^{2})} \frac{1}{1 + t ^{2}} d t

arctan (sin (x^{2})) - arctan (2)

Solution steps

\int_{2}^{s i n (x^{2})} \frac{1}{1 + t ^{2}} d t

Use the common integral:

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

= [arctan (t)]_{2}^{s i n (x^{2})}

Compute the boundaries:

arctan (sin (x^{2})) - arctan (2)

= arctan (sin (x^{2})) - arctan (2)

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