integral of 8sin(t)cos(t)

解答

\int 8 sin (t) cos (t) d t

4 sin^{2} (t) + C

求解步骤

\int 8 sin (t) cos (t) d t

提出常数:

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

= 8 \cdot \int sin (t) cos (t) d t

使用换元积分法

= 8 \cdot \int u d u

使用幂法则

= 8 \cdot \frac{u ^{2}}{2}

u = sin (t)

代回

= 8 \cdot \frac{sin ^{2} ( t )}{2}

化简

8 \cdot \frac{sin ^{2} ( t )}{2} : 4 sin^{2} (t)

= 4 sin^{2} (t)

解答补常数

= 4 sin^{2} (t) + C

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