Question

40/HT

Cho hàm số f(x) có f(0) = 0 và f'(x) = sin⁴x. Tính tích phân \(\int_0^{\frac{\pi}{2}}f\left(x\right)dx\)

Answer 1

\(f\left(x\right)=\int sin^4xdx=\int\left(\frac{1}{2}-\frac{1}{2}cos2x\right)^2dx\)

\(=\frac{1}{4}\int\left(1-2cos2x+cos^22x\right)dx=\frac{1}{4}\int\left(\frac{3}{2}-2cos2x+\frac{1}{2}cos4x\right)dx\)

\(=\frac{1}{4}\left(\frac{3}{2}x-sin2x+\frac{1}{8}sin4x\right)+C\)

\(f\left(0\right)=0\Rightarrow\frac{1}{4}\left(0-0+0\right)+C=0\Rightarrow C=0\)

\(\Rightarrow\int\limits^{\frac{\pi}{2}}_0f\left(x\right)dx=\frac{1}{4}\int\limits^{\frac{\pi}{2}}_0\left(\frac{3}{2}x-sin2x+\frac{1}{8}sin4x\right)dx\)

\(=\frac{1}{4}\left(\frac{3}{4}x^2+\frac{1}{2}cos2x-\frac{1}{32}cos4x\right)|^{\frac{\pi}{2}}_0\)

\(=\frac{3\pi^2-16}{64}\)