integral from 2x to x^2 of cos(t)sin(t)

解答

\int_{2 x}^{x^{2}} cos (t) sin (t) d t

\frac{sin ^{2} ( x ^{2} ) - sin ^{2} ( 2 x )}{2}

求解步骤

\int_{2 x}^{x^{2}} cos (t) sin (t) d t

使用换元积分法

= \int_{s i n (2 x)}^{s i n (x^{2})} u d u

使用幂法则

= [\frac{u ^{2}}{2}]_{s i n (2 x)}^{s i n (x^{2})}

带入上下限计算后得:

\frac{sin ^{2} ( x ^{2} )}{2} - \frac{sin ^{2} ( 2 x )}{2}

= \frac{sin ^{2} ( x ^{2} )}{2} - \frac{sin ^{2} ( 2 x )}{2}

化简

= \frac{sin ^{2} ( x ^{2} ) - sin ^{2} ( 2 x )}{2}

\int_{0}^{1 - x} \frac{( 1 - x - y ) ^{2}}{2} d y

\int_{0}^{2} x^{2} + x + 1 d x

\int_{2}^{3} \frac{1}{x ln ( x )} d x

\int_{0}^{1} 3 x (x^{2} + 1)^{3} d x

\int_{0}^{x} (x - t) f (t) d t