integral of 4tan^2(θ)

解答

\int 4 tan^{2} (θ) d θ

4 (- θ + tan (θ)) + C

求解步骤

\int 4 tan^{2} (θ) d θ

提出常数:

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

= 4 \cdot \int tan^{2} (θ) d θ

使用三角恒等式改写

= 4 \cdot \int - 1 + sec^{2} (θ) d θ

使用积分加法定则:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 4 (- \int 1 d θ + \int sec^{2} (θ) d θ)

\int 1 d θ = θ

\int sec^{2} (θ) d θ = tan (θ)

= 4 (- θ + tan (θ))

解答补常数

= 4 (- θ + tan (θ)) + C

\frac{\partial}{\partial x} (sin (3 x^{3} y - 5 x y^{3}))

\int cos^{7} (7 - x) sin (7 - x) d x

d er i v a t i v e (5 t + e^{t}) (3 - t)

(x^{2} y^{3} - \frac{1}{1 + 9 x ^{2}}) (\frac{d x}{d y}) + x^{3} y^{2} = 0

\int x^{2} 1 + x^{2} d x