int sin2x*e^(cos^(2)x)dx 01:41 int sin2x*e^(x)*dx 03:22 Evaluate : int(e^xdot(2+sin2x)/(2cos^2x)\ dx 02:54 Evaluate: int(1-sin2x)/(x+cos^2x)dx 03:01 int(1)/(cos^(2)x+sin2x)dx 02:13 int(sin2x)/((sin x+cos x)^(2))dx 07:44 int(dx)/(cos^(2)x+si...
int e^(sinx).sin2x dx = 02:42 int ("logx")^(2) dx = 04:31 int Cos^(-1)(2x^(2)-1)dx= 03:16 int cosh^(-1) ((x)/(3)) dx = 01:09 Find int(xsin^(-1)x)/(sqrt(1-x^(2)))dx 04:26 If int (x^(2) tan^(-1) x)/(1 + x^(2)) dx = xtan^(-1) x -...
$\int e^{2x} \cos x dx=$ A. \[\frac{{{e}^{x}}\,\left( \sin{(x)}+2\cos{(x)}\right) }{5}+C\] B. \[\frac{{{e}^{2x}}\,\left( \sin{(x)}+\cos{(x)}\right) }{5}+C\] C. \[\frac{{{e}^{2x}}\,\left( 2\sin{(x)}+2\cos{(x)}\right) }{5}+C...
Use the formula {eq}\int udv = uv - \int v du{/eq} and choose {eq}u = e^{2x}{/eq} to evaluate the next integral: {eq}\int e^{2x} sin (x)dx{/eq}. Indefinite Integral: The indefinite integral is written as: {eq}\int f(x)\ dx = F...
{eq}\int\frac{e^x}{\sqrt{1 - e^{2x}}} dx {/eq} Exponential and Inverse Sine Function: In the simplification of the given indefinite integral, we need to know about the first derivative of an exponential function (used in the substitution method) and the common integral value ...
Answer to: evaluate the integral \int e^x (e^x +1)^2 dx By signing up, you'll get thousands of step-by-step solutions to your homework questions...
答案 解析 null 本题来源 题目:计算下列不定积分:(1) $\int (3x^2 2x 1) \, dx$(2) $\int \frac{1}{x \sqrt{x^2 1}} \, dx$(3) $\int e^x \cdot \sin x \, dx$ 来源: 高数上练习题 收藏 反馈 分享
12.已知f1(x)=e-x+sinx,fn+1(x)是fn(x)的导函数,即f2(x)=f1′(x),f3(x)=f2′(x),…,fn+1(x)=fn′(x),n∈N*,则f2016(x)=( ) A. e-x+sinx B. -e-x+cosx C. e-x-sinx D. -e-x-cosx 查看答案和解析>> 同步练习册答案全...
Evaluate the triple integral \int_{-1}^{2} \int_{-\frac{x^2}{2^{\frac{x^2}{2 \int_{x^2 + 3y^2}^{8 - x^2 -y^2}xz dz dy dx. Calculate the double integral \int\int_Rx\cos(2x+y)\ dA where R is the region \displaystyle 0\leq...
12.已知函数y=f(x-1)定义域是[-2,3],则y=f(2x+1)的定义域是( ) A.[−2,12][−2,12]B.[-1,4]C.[−52,52][−52,52]D.[-3,7] 科目:来源:题型:填空题 11.若函数y=sin(ωx+5π35π3)(ω>0)的最小正周期是π3π3,则ω=6. ...