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受欢迎的微积分 >

integral of tan(x)sqrt(1+sec^4(x))

解答

\int tan (x) 1 + sec^{4} (x) d x

解答

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

求解步骤

\int tan (x) 1 + sec^{4} (x) d x

使用三角换元法

= \int - \frac{u ^{4} + 1}{u ^{3}} d u

提出常数:

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

= - \int \frac{u ^{4} + 1}{u ^{3}} d u

使用分布积分法

= - (- \frac{u ^{4} + 1}{2 u ^{2}} - \int - \frac{u}{u ^{4} + 1} d u)

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

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

u = cos (x)

代回

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

化简

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

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

解答补常数

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

作图

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