Evaluating the Integral of x arctan(x^2) Using Techniques of Integration

Integration is a fundamental concept in calculus, and evaluating specific integrals requires a deep understanding of various integration techniques. One such integral involves the arctan function, which can be tackled using integration by parts. This article will guide you through the process of evaluating the integral (I int_0^1 xarctan(x^2) dx), showcasing the application of integration by parts and other related techniques.

Introduction to the Integral

Let's consider the integral (I int_0^1 xarctan(x^2) dx). This integral stands as a challenging problem, as it combines the arctan function with a linear function multiplied by the variable of integration. To solve this, we will employ a clever substitution and multiple applications of integration by parts.

Substitution and Simplification

To begin, we make the substitution (t arctan(x)), which implies (x tan(t)). The limits of integration change to (t in [0, frac{pi}{4}]). Substituting these into the integral, we get:

[I int_0^{pi/4} tan(t) cdot t^2 cdot sec^2(t) dt int_0^{pi/4} t^2 cdot tan(t) sec^2(t) dt]

First Integration by Parts

We now apply integration by parts, with:

(u t^2) (dv tan(t) sec^2(t) dt)

The next step involves computing (du) and (v) as follows:

(du 2t dt) (v frac{1}{2} tan^2(t))

Substituting (u, v, du,) and (dv) into the integration by parts formula:

[I left[frac{1}{2} t^2 tan^2(t)right]_0^{pi/4} - int_0^{pi/4} t cdot tan^2(t) dt]

Evaluating the first term, we obtain:

[left[frac{1}{2} t^2 tan^2(t)right]_0^{pi/4} frac{1}{2} left(frac{pi}{4}right)^2 frac{pi^2}{32}]

The remaining integral can be decomposed as follows:

[int_0^{pi/4} t cdot tan^2(t) dt int_0^{pi/4} t cdot (sec^2(t) - 1) dt int_0^{pi/4} t cdot sec^2(t) dt - int_0^{pi/4} t dt]

This simplifies to:

[I frac{pi^2}{32} - int_0^{pi/4} t sec^2(t) dt - left[frac{1}{2} t^2right]_0^{pi/4}]

Evaluating these integrals, we get:

[I frac{pi^2}{32} - left[frac{1}{2} t^2right]_0^{pi/4} - int_0^{pi/4} t sec^2(t) dt frac{pi^2}{32} - frac{pi^2}{32} - int_0^{pi/4} t sec^2(t) dt]

Hence:

[I frac{pi^2}{16} - int_0^{pi/4} t sec^2(t) dt]

Second Integration by Parts

For the remaining integral (int_0^{pi/4} t sec^2(t) dt), we apply integration by parts once more, with:

(u t) (dv sec^2(t) dt)

Then:

(du dt) (v tan(t))

Substituting, we get:

[ int_0^{pi/4} t sec^2(t) dt left[t tan(t)right]_0^{pi/4} - int_0^{pi/4} tan(t) dt]

Evaluating the first term and computing the second:

[left[t tan(t)right]_0^{pi/4} frac{pi}{4}]

[int_0^{pi/4} tan(t) dt left[-ln|cos(t)|right]_0^{pi/4} -lnleft(frac{1}{sqrt{2}}right) ln(1) -lnleft(frac{1}{2}right) ln(2)]

Bringing it all together:

[int_0^{pi/4} t sec^2(t) dt frac{pi}{4} ln(2)]

Substituting this back into our expression for (I) gives:

[I frac{pi^2}{16} - left(frac{pi}{4} ln(2)right) frac{pi^2}{16} - frac{pi}{4} - frac{ln(2)}{2}]

Conclusion

We have thus evaluated the integral (I int_0^1 xarctan(x^2) dx), concluding that:

[int_0^1 x arctan(x^2) dx frac{pi^2}{16} - frac{pi}{4} - frac{ln(2)}{2}]

This result showcases the power of integration by parts and substitution in tackling complex integrals involving the arctan function. By breaking down the integral into manageable parts and applying these techniques, we can solve even seemingly daunting problems in calculus.