integration - Find the value of the integral: $int_{0}^{frac{pi}{4}} frac{1}{3cos^2theta + sin^2theta}dtheta$

Using the substitution $t = \tan\theta$ , find the value of the integral:

$$\int_{0}^{\pi/4} \cfrac{1}{3\cos^2\theta + \sin^2\theta}d\theta$$

I isolated $\theta$ in $t = \tan\theta$, and then substituted the expression into the integral. I then replaced $d\theta$ by $dt$, which resulted in a very ugly, large expression.

The following is the answer given:

I do not understand how the first integral was obtained.

Answer

Draw a right triangle with an angle $\theta$. Since $\tan\theta = t$, you can label the side opposite $\theta$ as $t$, and the side adjacent to $\theta$ as $1$. Then the hypotenuse is $\sqrt{t^2+1}$. Therefore:

$$\cos\theta = \frac{1}{\sqrt{t^2+1}} \qquad \text{and} \qquad \sin\theta = \frac{t}{\sqrt{t^2+1}}$$

Can you take it from here?