Question

tìm nguyên hàm F(x) của hàm số f(x) thỏa mãn điều kiện tương ứng

a) \(f\left(x\right)=tan^2x\) và \(F\left(\dfrac{\pi}{4}\right)=-2\)

b) \(f\left(x\right)=cot^2x-2\) và \(F\left(\dfrac{\pi}{4}\right)=15\)

Answer 1

a: \(F\left(x\right)=\int tan^2x.dx=\int\left(tan^2x+1-1\right)dx\)

\(=tanx-x+C\)

\(F\left(\dfrac{\Omega}{4}\right)=-2\)

=>\(tan\left(\dfrac{\Omega}{4}\right)-\dfrac{\Omega}{4}+C=2\)

=>\(C=1+\dfrac{\Omega}{4}\)

=>\(F\left(x\right)=tanx-x+1+\dfrac{\Omega}{4}\)

b: \(F\left(x\right)=\int f\left(x\right)=\int\left(cot^2x-2\right)dx=\int\left(cot^2x+1-3\right)dx\)

=-cotx-3x+C

\(F\left(\dfrac{\Omega}{4}\right)=15\)

=>\(-cot\left(\dfrac{\Omega}{4}\right)-3\cdot\dfrac{\Omega}{4}+C=15\)

=>\(-1-\dfrac{3\Omega}{4}+C=15\)

=>\(C=16+\dfrac{3\Omega}{4}\)

=>\(F\left(x\right)=-cotx-3x+16+\dfrac{3\Omega}{4}\)