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人気のある微分積分 >

integral of sec^2(xta)n^2x

解

\int sec^{2} (x t a) n^{2} x d x

解

n^{2} \frac{1}{a ^{2} t ^{2}} (a t x tan (a t x) + ln ∣ cos (a t x) ∣) + C

解答ステップ

\int sec^{2} (x t a) n^{2} x d x

定数を除く:

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

= n^{2} \cdot \int sec^{2} (a t x) x d x

u置換積分法を適用する

= n^{2} \cdot \int \frac{u sec ^{2} ( u )}{a ^{2} t ^{2}} d u

定数を除く:

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

= n^{2} \frac{1}{a ^{2} t ^{2}} \cdot \int u sec^{2} (u) d u

部分積分を適用する

= n^{2} \frac{1}{a ^{2} t ^{2}} (u tan (u) - \int tan (u) d u)

\int tan (u) d u = - ln ∣ cos (u) ∣

= n^{2} \frac{1}{a ^{2} t ^{2}} (u tan (u) - (- ln ∣ cos (u) ∣))

代用を戻す

u = a t x

= n^{2} \frac{1}{a ^{2} t ^{2}} (a t x tan (a t x) - (- ln ∣ cos (a t x) ∣))

簡素化

= n^{2} \frac{1}{a ^{2} t ^{2}} (a t x tan (a t x) + ln ∣ cos (a t x) ∣)

定数を解答に追加する

= n^{2} \frac{1}{a ^{2} t ^{2}} (a t x tan (a t x) + ln ∣ cos (a t x) ∣) + C

グラフ

人気の例

tangent f(x)=x^3-3/2 x^2+x,\at x=1

t an g e n t f (x) = x^{3} - \frac{3}{2} x^{2} + x, at x = 1

sum from n=2 to infinity of ((-1)^n)/(sqrt(n)ln(n))

n = 2 \sum \infty \frac{( - 1 ) ^{n}}{n ln ( n )}

integral of sqrt(4+y^2)

\int 4 + y^{2} d y

integral of (14)/((x-1)^2(x-2)^2)

\int \frac{14}{( x - 1 ) ^{2} ( x - 2 ) ^{2}} d x

f(x)=arcsin(4x)

f (x) = arcsin (4 x)