$$\int_0^{1} \frac{1}{1+\tan^2{x}}\,\mathrm{d}x$$
$$\int_0^{1} \frac{1}{1+a^2\tan^2{x}}\,\mathrm{d}x $$
How can I evaluate these integrals? I don't know which substitution to use to solve them.
I wanted to use the solution of the second integral to evaluate this integral: $$I (1)=\int_0^{1} \frac{x}{tan{x}}\,\mathrm{d}x $$ by converting it to the integral with the parameter as I obtain: $$I'(a)=\int_0^{1} \frac{1}{1+a^2\tan^2{x}}\,\mathrm{d}x $$
HINT: For the first one, you don't even need to use substitution. Use the trigonometric identity $$1+\tan^2 \theta=\sec^2 \theta$$ and the integral will become much easier.
Now for the second one. Start off with $$\int \frac{dx}{1+a^2\tan^2 x}$$ and make the substitution $x \to \arctan u$. You will end up with $$\int \frac{du}{(1+u^2)(1+a^2u^2)}$$ Now, using partial fractions, this is equal to $$\frac{1}{a^2-1}\int \bigg(\frac{a^2}{1+a^2u^2}-\frac{1}{1+u^2}\bigg)du$$ Now, using the fact that $$\frac{d}{d\phi} \arctan\phi=\frac{1}{1+\phi^2}$$ You should be able to easily finish this integral.