Question

\(\int_0^{\dfrac{\pi}{4}}\) \(\dfrac{3sin^2x-4cos^2x}{cos^2x}dx\)

Answer 1

\(=\int\limits^{\dfrac{\pi}{4}}_0\dfrac{3\left(1-cos^2x\right)-4cos^2x}{cos^2x}dx=\int\limits^{\dfrac{\pi}{4}}_0\dfrac{3-7cos^2x}{cos^2x}dx\)

\(=\int\limits^{\dfrac{\pi}{4}}_0\left(\dfrac{3}{cos^2x}-7\right)dx=\left(3tanx-7x\right)|^{\dfrac{\pi}{4}}_0=...\)