integral of (sin^5(x)-cos^2(x))/(cos(x))

解答

\int \frac{sin ^{5} ( x ) - cos ^{2} ( x )}{cos ( x )} d x

- \frac{1}{2} ln (cos^{2} (x)) - \frac{1}{2} sin^{2} (x) - \frac{1}{4} sin^{4} (x) - sin (x) + C

求解步骤

\int \frac{sin ^{5} ( x ) - cos ^{2} ( x )}{cos ( x )} d x

使用三角换元法

= \int \frac{u ^{5} - 1 + u ^{2}}{1 - u ^{2}} d u

乘开

\frac{u ^{5} - 1 + u ^{2}}{1 - u ^{2}} : \frac{u ^{5}}{1 - u ^{2}} - \frac{1}{1 - u ^{2}} + \frac{u ^{2}}{1 - u ^{2}}

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 ^{5}}{1 - u ^{2}} d u - \int \frac{1}{1 - u ^{2}} d u + \int \frac{u ^{2}}{1 - u ^{2}} d u

\int \frac{u ^{5}}{1 - u ^{2}} d u = - \frac{u ^{4}}{4} - \frac{u ^{2}}{2} - \frac{1}{2} ln - u^{2} + 1

\int \frac{1}{1 - u ^{2}} d u = \frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2}

\int \frac{u ^{2}}{1 - u ^{2}} d u = \frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2} - u

= - \frac{u ^{4}}{4} - \frac{u ^{2}}{2} - \frac{1}{2} ln - u^{2} + 1 - (\frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2}) + \frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2} - u

u = sin (x)

代回

= - \frac{sin ^{4} ( x )}{4} - \frac{sin ^{2} ( x )}{2} - \frac{1}{2} ln - sin^{2} (x) + 1 - (\frac{ln ∣ sin ( x ) + 1 ∣}{2} - \frac{ln ∣ sin ( x ) - 1 ∣}{2}) + \frac{ln ∣ sin ( x ) + 1 ∣}{2} - \frac{ln ∣ sin ( x ) - 1 ∣}{2} - sin (x)

化简

- \frac{sin ^{4} ( x )}{4} - \frac{sin ^{2} ( x )}{2} - \frac{1}{2} ln - sin^{2} (x) + 1 - (\frac{ln ∣ sin ( x ) + 1 ∣}{2} - \frac{ln ∣ sin ( x ) - 1 ∣}{2}) + \frac{ln ∣ sin ( x ) + 1 ∣}{2} - \frac{ln ∣ sin ( x ) - 1 ∣}{2} - sin (x) : - \frac{1}{2} ln (cos^{2} (x)) - \frac{1}{2} sin^{2} (x) - \frac{1}{4} sin^{4} (x) - sin (x)

= - \frac{1}{2} ln (cos^{2} (x)) - \frac{1}{2} sin^{2} (x) - \frac{1}{4} sin^{4} (x) - sin (x)

解答补常数

= - \frac{1}{2} ln (cos^{2} (x)) - \frac{1}{2} sin^{2} (x) - \frac{1}{4} sin^{4} (x) - sin (x) + C

\int 2 x^{3} (ln (x^{2})) d x

\int (8 x^{9} - 3 x^{6} + 12 x^{3}) d x

\int (4 x^{2} + x^{- 3}) d x

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

\int (x + 4)^{7} d x