integral of sec^{13}(2θ)tan^3(2θ)

解答

\int sec^{13} (2 θ) tan^{3} (2 θ) d θ

\frac{sec ^{15} ( 2 θ )}{30} - \frac{sec ^{13} ( 2 θ )}{26} + C

求解步骤

\int sec^{13} (2 θ) tan^{3} (2 θ) d θ

使用三角恒等式改写

= \int (- 1 + sec^{2} (2 θ)) tan (2 θ) sec^{13} (2 θ) d θ

使用换元积分法

= \int \frac{u ^{12} ( u ^{2} - 1 )}{2} d u

乘开

\frac{u ^{12} ( u ^{2} - 1 )}{2} : \frac{u ^{14}}{2} - \frac{u ^{12}}{2}

= \int \frac{u ^{14}}{2} - \frac{u ^{12}}{2} d u

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

= \int \frac{u ^{14}}{2} d u - \int \frac{u ^{12}}{2} d u

\int \frac{u ^{14}}{2} d u = \frac{u ^{15}}{30}

\int \frac{u ^{12}}{2} d u = \frac{u ^{13}}{26}

= \frac{u ^{15}}{30} - \frac{u ^{13}}{26}

u = sec (2 θ)

代回

= \frac{sec ^{15} ( 2 θ )}{30} - \frac{sec ^{13} ( 2 θ )}{26}

解答补常数

= \frac{sec ^{15} ( 2 θ )}{30} - \frac{sec ^{13} ( 2 θ )}{26} + C

\int 5 (x^{4} + 5)^{5} + C d x

\int (6 x^{3} + 1) e^{3 x^{4} + 2 x} d x

\int \frac{x ^{5}}{x ^{2} + 16} d x

\int \frac{1}{4 x ^{2} + 12 x + 5} d x

\int sec^{3} (5 x) tan (5 x) d x