integral of sin(t)*sin(t)

解答

\int sin (t) \cdot sin (t) d t

\frac{1}{2} (t - \frac{1}{2} sin (2 t)) + C

求解步骤

\int sin (t) sin (t) d t

化简

sin (t) sin (t) : sin^{2} (t)

= \int sin^{2} (t) d t

使用三角恒等式改写

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

提出常数:

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

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

使用积分加法定则:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

\int 1 d t = t

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

= \frac{1}{2} (t - \frac{1}{2} sin (2 t))

解答补常数

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

n \to \infty lim (7 n^{\frac{5}{4}})

y^{^{'}} - y = 9 t e^{2} t, y (0) = 1

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

\int_{0}^{\frac{π}{4}} cos^{4} (x) d x

d er i v a t i v e 25 x^{4} + 30 x^{3} - 11 x^{2} - 12 x + 4