What is the antiderivative of cosx. Again, people memorize that the antiderivative of cosx is sinx. What's the antiderivative of tan? tan x = - ln|cos x| + C. What are the applications of derivatives? Applications of Derivatives in Maths Finding Rate of Change of a Quantity. Finding the...
Use integration by parts to find the integral of:[Hint: In (7) write lnx as 1lnx and in (9) write arctanxdn as 1arctanax.](1)xe^x(2)xsinx(3)x^2lnx(4)xsin3x(5)xcos2x(6)xsec^2x(7)lnx(8)(lnx)^2(9)arctanxdn ...
Find the integral of 2cos^5(2x) sin(2x) dx. Integrate (1 - sinx)/(cosx) Integrate \sqrt{(2((sin(x))^2 + 50((cos(x))^2 - 10 sin(x)cos(x))} Integrate the function: cos x/sqrt(4 - sin^2 x) Integrate by any method and show the steps. ((sin x + cos x)/ sin...
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 of the integral of sec^3(1 - 2x) dx? What is the antiderivative of the integral of (5x + 3)/(x^3 - 2x^2 - 3x) dx?
求不定积分Integral of (a*sinx + b*cosx)/(c*sinx + d*cosx) dx(sinx)'=cosx (cosx)'=-sinx,在导数里,是2个非常友好的函数,本题的分子分母结构一样,只是系数不同,可以利用这个条件,用待定系数的方法,将被积函数分解成2个容易积分的函数,求解出答案。, 视频播放量
67. Show that the average value of sin2tsin2t over [0,2π][0,2π] is equal to 1212. Without further calculation, determine whether the average value of sin2tsin2t over [0,π][0,π] is also equal to 1212.68. Show that the average value of cos2tcos2t over [0,2...
Evaluate the indefinite integral. \int \frac{cosx}{sin^2x} dx Evaluate the indefinite integral. \int_0^\frac{1}{2} \frac{x^2}{(1-x^2)^\frac{1}{2 Evaluate the indefinite integral ((1+x)/(1+x^2))dx Evaluate the indefinite integral. \int \frac{2x}{x^4 + 1} dx ...
Evaluate: the integral from pi/2 to pi the integral from 0 to x^2 of 1/x cos y/x dydx. Evaluate the integral 1. integral from 0 to 2 (dt /(squarerroot 4 + t^2)) 2. integral from 0 to pi/2 cosx/(squqreroot(1+sin^2x)) ...
Consider the integral ∫2π0dx5−2cosx making the substitution tan(x2)=t, we have I=∫2π0dx5−2cosx =∫002dt(1+t2)[5−21−t21+t2]=0 The result is obviously wrong, since the integrand is positive and consequently the integral of this function cannot be equal to zero. Find ...
2x+3dxAns: 1 10 (2x+3) 5/2 − 1 2 (2x+3) 3/2 +C B−2 Productsofsinesandcosines B39. 2cosxsinxdxAns:sin 2 x+C B40. 2sin3xcos2xdxAns:− 1 5 cos5x−cosx+C B41. 14cos3xcos4xdxAns:sin7x+7sinx+C B42. 8sin3xsinxdxAns:−sin4x+2sin2x+C B43. 4sin 2 xdx...