For problem 5: Find the series for arctan(x), which you find by integrating the series for 1/(1+x^2)=1-x^2+x^4-...

The series for arctan(x) is x-x^3/3+x^5/5-...

The series for arctan(x^2) is x^2-x^6/3+x^10/5-...

The interval of convergence is (-1,1), same as for the original geometric series.

For problem 6: Take the series for sin(t)=t-t^3/3!+t^5/5!-...

Divide through by t: 1-t^2/3!+t^4/5!-...

And integrate and plug in the upper limit (which should have been x): x-x^3/(3*3!)+x^5/(5*5!)-...

Since this is alternating, you only need take the first three terms of the series to get the desired accuracy (three decimal places). In general, for an alternating series, you need only take terms until the absolute value of the terms falls below the desired accuracy.

Please stop by for any additional questions tomorrow morning. -Prof. B^2