كيف يمكنك دمج int sqrt (3 (1-x ^ 2)) dx باستخدام استبدال المثلثية؟

إجابة:

#int sqrt (3 (1-x ^ 2)) dx = sqrt3 / 4sin2theta + sqrt3 / 2 theta + C #

تفسير:

# x = sintheta ، dx = cos theta d theta #

#intsqrt (3 (1-sin ^ 2theta)) * cos theta d theta = intsqrt (3 (cos ^ 2theta)) cos theta d theta #

# = intsqrt3 cos theta cos theta d theta #

# = sqrt 3intcos ^ 2 theta d theta #

# = sqrt3 int1 / 2 (cos2 theta + 1) d theta #

# = sqrt3 / 2 int (cos2 theta + 1) d theta #

# = sqrt3 / 2 1/2 sin2theta + theta #

# = sqrt3 / 4sin2theta + sqrt3 / 2 theta + C #