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受欢迎的微积分 >

integral of tan^2(xse)c^2x

解答

\int tan^{2} (x se) c^{2} x d x

解答

\frac{c ^{2}}{2 e ^{2} s ^{2}} (- e^{2} s^{2} x^{2} + 2 es x tan (es x) + 2 ln ∣ cos (es x) ∣) + C

求解步骤

\int tan^{2} (x se) c^{2} x d x

提出常数:

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

= c^{2} \cdot \int tan^{2} (se x) x d x

使用换元积分法

= c^{2} \cdot \int \frac{u tan ^{2} ( u )}{e ^{2} s ^{2}} d u

提出常数:

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

= c^{2} \frac{1}{e ^{2} s ^{2}} \cdot \int u tan^{2} (u) d u

使用分布积分法

= c^{2} \frac{1}{e ^{2} s ^{2}} (u (- u + tan (u)) - \int - u + tan (u) d u)

\int - u + tan (u) d u = - \frac{u ^{2}}{2} - ln ∣ cos (u) ∣

= c^{2} \frac{1}{e ^{2} s ^{2}} (u (- u + tan (u)) - (- \frac{u ^{2}}{2} - ln ∣ cos (u) ∣))

u = se x

代回

= c^{2} \frac{1}{e ^{2} s ^{2}} (se x (- se x + tan (se x)) - (- \frac{( se x ) ^{2}}{2} - ln ∣ cos (se x) ∣))

化简

c^{2} \frac{1}{e ^{2} s ^{2}} (se x (- se x + tan (se x)) - (- \frac{( se x ) ^{2}}{2} - ln ∣ cos (se x) ∣)) : \frac{c ^{2}}{2 e ^{2} s ^{2}} (- e^{2} s^{2} x^{2} + 2 es x tan (es x) + 2 ln ∣ cos (es x) ∣)

= \frac{c ^{2}}{2 e ^{2} s ^{2}} (- e^{2} s^{2} x^{2} + 2 es x tan (es x) + 2 ln ∣ cos (es x) ∣)

解答补常数

= \frac{c ^{2}}{2 e ^{2} s ^{2}} (- e^{2} s^{2} x^{2} + 2 es x tan (es x) + 2 ln ∣ cos (es x) ∣) + C

作图

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