解答
∫tan(x)sec2(x)dx
解答
−22(−81(ln2cot(x)+22cot(x)+2−2arctan(2cot(x)+1))+81(ln2cot(x)−22cot(x)+2+2arctan(2cot(x)−1)))+xtan23(x)−arctan(tan(x))tan23(x)−221ln2tan(x)+22tan(x)+2−21arctan(2tan(x)+1)+221ln2tan(x)−22tan(x)+2−21arctan(2tan(x)−1)+2tan(x)+C
求解步骤
∫tan(x)sec2(x)dx
使用三角恒等式改写
=∫tan(x)1+tan2(x)dx
乘开 tan(x)1+tan2(x):tan(x)1+tan23(x)
=∫tan(x)1+tan23(x)dx
使用积分加法定则: ∫f(x)±g(x)dx=∫f(x)dx±∫g(x)dx=∫tan(x)1dx+∫tan23(x)dx
∫tan(x)1dx=−22(−81(ln2cot(x)+22cot(x)+2−2arctan(2cot(x)+1))+81(ln2cot(x)−22cot(x)+2+2arctan(2cot(x)−1)))
∫tan23(x)dx=xtan23(x)−arctan(tan(x))tan23(x)−221ln2tan(x)+22tan(x)+2−21arctan(2tan(x)+1)+221ln2tan(x)−22tan(x)+2−21arctan(2tan(x)−1)+2tan(x)
=−22(−81(ln2cot(x)+22cot(x)+2−2arctan(2cot(x)+1))+81(ln2cot(x)−22cot(x)+2+2arctan(2cot(x)−1)))+xtan23(x)−arctan(tan(x))tan23(x)−221ln2tan(x)+22tan(x)+2−21arctan(2tan(x)+1)+221ln2tan(x)−22tan(x)+2−21arctan(2tan(x)−1)+2tan(x)
解答补常数=−22(−81(ln2cot(x)+22cot(x)+2−2arctan(2cot(x)+1))+81(ln2cot(x)−22cot(x)+2+2arctan(2cot(x)−1)))+xtan23(x)−arctan(tan(x))tan23(x)−221ln2tan(x)+22tan(x)+2−21arctan(2tan(x)+1)+221ln2tan(x)−22tan(x)+2−21arctan(2tan(x)−1)+2tan(x)+C