integral from 2x^2 to cos(x) of ln(5t-2)

解

\int_{2 x^{2}}^{c o s (x)} ln (5 t - 2) d t

cos (x) ln (5 cos (x) - 2) - cos (x) - \frac{2}{5} ln (5 cos (x) - 2) + 2 x^{2} + \frac{2}{5} ln (10 x^{2} - 2) - 2 x^{2} ln (10 x^{2} - 2)

解答ステップ

\int_{2 x^{2}}^{c o s (x)} ln (5 t - 2) d t

u置換積分法を適用する

= \int_{10 x^{2} - 2}^{5 c o s (x) - 2} ln (u) \frac{1}{5} d u

定数を除く:

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

= \frac{1}{5} \cdot \int_{10 x^{2} - 2}^{5 c o s (x) - 2} ln (u) d u

部分積分を適用する

= \frac{1}{5} [u ln (u) - \int 1 d u]_{10 x^{2} - 2}^{5 c o s (x) - 2}

\int 1 d u = u

= \frac{1}{5} [u ln (u) - u]_{10 x^{2} - 2}^{5 c o s (x) - 2}

限度を計算する:

5 cos (x) ln (5 cos (x) - 2) - 5 cos (x) - 2 ln (5 cos (x) - 2) + 10 x^{2} + 2 ln (10 x^{2} - 2) - 10 x^{2} ln (10 x^{2} - 2)

= \frac{1}{5} (5 cos (x) ln (5 cos (x) - 2) - 5 cos (x) - 2 ln (5 cos (x) - 2) + 10 x^{2} + 2 ln (10 x^{2} - 2) - 10 x^{2} ln (10 x^{2} - 2))

簡素化

= cos (x) ln (5 cos (x) - 2) - cos (x) - \frac{2}{5} ln (5 cos (x) - 2) + 2 x^{2} + \frac{2}{5} ln (10 x^{2} - 2) - 2 x^{2} ln (10 x^{2} - 2)