Evaluation of $1-\frac{1}{7}+\frac{1}{9}-\frac{1}{15}+\frac{1}{17}-\dotsb$ 
How can we calculate sum of following infinite  series
$\displaystyle \bullet\; 1+\frac{1}{3}-\frac{1}{5}-\frac{1}{7}+\frac{1}{9}+\frac{1}{11}-\cdots$
$\displaystyle \bullet\; 1-\frac{1}{7}+\frac{1}{9}-\frac{1}{15}+\frac{1}{17}-\cdots$

$\textbf{My Try:}$ Let $$S = \int_0^1 (1+x^2-x^4-x^6+x^8+x^{10}+\cdots) \, dx$$ 
So we get $$S=\int_0^1 \left(1-x^4+x^8-\cdots\right)dx+x^2\int_0^1 (1-x^4+x^8-\cdots)\,dx$$
So we get $$S=\int_0^1 \frac{1+x^2}{1+x^4} \, dx= \frac{\pi}{2\sqrt{2}}$$ after that we can solve it
Again for second One, Let $$S=\int_0^1 (1-x^6+x^8-x^{14}+x^{16}+\cdots)$$
So we get $$S=\int_0^1 (1+x^8+x^{16}+\cdots) \, dx-\int_0^1 (x^6+x^{14}+
\cdots)\,dx$$
So we get $$S=\int_{0}^{1}\frac{1-x^6}{1-x^8}dx = \int_{0}^{1}\frac{x^4+x^2+1}{(x^2+1)(x^4+1)}dx$$
Now how can i solve after that, Help me
Thanks 

For the second, you're doing fine. You need to use partial fractions for the integral you've gotten. $x^2 + 1$ is irreducible, but the $x^4 + 1$ may be giving you troubles. 
Hint: 
$$x^4 + 1 = 
(x^2 + \sqrt{2}x + 1)(x^2 - \sqrt{2}x + 1)
$$
