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受欢迎的微积分 >

integral of (4sin(2x))/(1+cos^4(x))

解答

\int \frac{4 sin ( 2 x )}{1 + cos ^{4} ( x )} d x

解答

- 4 arctan (cos^{2} (x)) + C

求解步骤

\int \frac{4 sin ( 2 x )}{1 + cos ^{4} ( x )} d x

提出常数:

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

= 4 \cdot \int \frac{sin ( 2 x )}{1 + cos ^{4} ( x )} d x

使用三角恒等式改写

= 4 \cdot \int \frac{2 cos ( x ) sin ( x )}{1 + cos ^{4} ( x )} d x

提出常数:

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

= 4 \cdot 2 \cdot \int \frac{cos ( x ) sin ( x )}{1 + cos ^{4} ( x )} d x

使用换元积分法

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

提出常数:

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

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

使用换元积分法

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

提出常数:

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

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

使用积分常见定则:

\int \frac{1}{1 + v ^{2}} d v = arctan (v)

= 4 \cdot 2 (- \frac{1}{2} arctan (v))

代回

= 4 \cdot 2 (- \frac{1}{2} arctan (cos^{2} (x)))

化简

4 \cdot 2 (- \frac{1}{2} arctan (cos^{2} (x))) : - 4 arctan (cos^{2} (x))

= - 4 arctan (cos^{2} (x))

解答补常数

= - 4 arctan (cos^{2} (x)) + C

作图

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