integral of-sin^2(x)+cos^2(x)

解答

\int - sin^{2} (x) + cos^{2} (x) d x

\frac{1}{2} sin (2 x) + C

求解步骤

\int - sin^{2} (x) + cos^{2} (x) d x

使用三角恒等式改写

= \int cos (2 x) d x

使用换元积分法

= \int cos (u) \frac{1}{2} d u

提出常数:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int cos (u) d u

使用积分常见定则:

\int cos (u) d u = sin (u)

= \frac{1}{2} sin (u)

u = 2 x

代回

= \frac{1}{2} sin (2 x)

解答补常数

= \frac{1}{2} sin (2 x) + C

\int_{1}^{3} \frac{x ^{2} - 1}{x} d x

\int arctan (x - 1) d x

\int \frac{1}{( x ^{3} - 1 )} d x

\int \frac{2 x ^{2} - 1}{x ( x + 1 ) ^{3}} d x

\frac{d}{d x} (tan (1 - x))