百度试题 结果1 题目Evaluate the given definite integral.x cos (2x=) dx 相关知识点: 试题来源: 解析 −(√2)/8−(√2)/8 反馈 收藏
For instance, it is known that {eq}\frac{d}{dx} \sin(x) = \cos(x) {/eq} and, therefore, one antiderivative of cos(x) is sin(x). A note to say that typically the general antiderivative involves a + c term at the end since the derivative of a constant is zero. The ...
2elpha2 第一个注意到分母可以变成【3+cos(2x)】/2而正好sin(2x)dx=-dcos(2x)/2题目就转化为-∫dcos(2x)/【3+cos(2x)】 2022-01-23 12:134回复 晓之车高山老师 其实up表达的意思就是,被积函数某个地方稍有改动,对应不定积分表达式就可能有很大的变化,甚至完全不同 2022-01-24 03:072回复 QNのstar...
\int^{r}_{o} x((\sqrt{r^{2} - x^{2) - (-\sqrt{r^{2} - x^{2)) dx What's the integral of sec(x)tan(x) dx? What is the integral of (e^{-2x})(cos x) dx ? What is the integral from 7 to 8 of \frac{(x)}{(x^2 + 6x + 13)}dx What is the antiderivative...
\int_1^2 {\left( {x + 1} \right)}^2 \over x}dx} Evaluate the integral: \int_0^1(1 + x)\;dx Evaluate the integral I=\int_0^1 2e^{x^2+ \ln x}dx Find the integral (dx) / (cos^2x sqrt (1+ tan x)). find the integral \int tan^6 x sec^2 xdx Find the ...
∫ sin x dx = – cos x + C ∫ cos x dx = sin x + C ∫ sec2x dx = tan x + C ∫ csc2x dx = -cot x + C ∫ sec x (tan x) dx = sec x + C ∫ csc x ( cot x) dx = – csc x + C ∫ (1/x) dx = ln |x| + C ...
I=int{0}^{pi / 2} cos ^{2} x d x =int{0}^{pi / 2} frac{1+cos 2 {x}}{2} {dx} ldots ldots ldots ldots . . . ldots(cos 2 {x}=2 cos ^{2} {x}-1) =frac{1}{2}[x+frac{sin 2 x}{2}]{0}^{pi / 2} =frac{1}{2}[(frac{pi}{2}+frac{sin pi}{2})...
Evaluate the integral: \ \int_0^{\frac{\pi}{3 (\cos t-\sec^2t \tan t)dt Evaluate the integral. integral^6_4 \frac{x}{x^2 + 4x + 13} dx Evaluate the integral: \int_2^{\infty} \frac{2}{(x + 3)^{\frac{3}{2}dx Evaluate the integral: \frac{x}{(1+(e^...
By using the properties of definite integrals, evaluate the integrals∫π20cos2xdx View Solution Evaluate the definite integrals∫π20cos2xdx View Solution Evaluate the following definite integral:∫π/20(a2cos2x+b2sin2x)dx View Solution
Integrate by parts using the formula(∫ udv=uv-∫ vdu), where ( u=x) and ( dv=(cos)(2x)). ( x(1/2(sin)(2x))]_0^((π )/4)-(∫ )_0^((π )/4)1/2(sin)(2x)dx) Simplify. ( (x(sin)(2x))/2]_0^((π )/4)-(∫ )_0^((π )/4)((sin)(2x))/2dx) Since ...