Calculate the Integral of … CLR+–×÷^√f(x)π() √3√4√n√ You can also input: •sqrt(…) •root(n, …) lnlog10lognexpexabs|x| sincostancscseccot arcsinsin-1arccoscos-1arctantan-1 arccsccsc-1arcsecsec-1arccotcot-1 ...
Evaluate {eq}\displaystyle \int \sin (2x)\cos (2x) \ dx {/eq} Integration by Substitution: Integration by substitution can be used when one part of our integral is the derivative of the other portion. We recall the substitution rule for integration: If {eq}u=g(x) {/eq} is a d...
Answer to: Integrate the following integral: integral of (-3sin x - 3sec x)/(tan x) dx. By signing up, you'll get thousands of step-by-step...
function to integrate: Variable 1: Variable 2: Also include:domains of integration for variables Compute More than just an online double integral solver Wolfram|Alpha is a great tool for calculating indefinite and definite double integrals. Compute volumes under surfaces, surface area and other types...
How do you integrate cos(x)? To integrate cos(x) we find the antiderivative of cos(x) using the Fundamental Theorem of Calculus and derivative rules. Since the antiderivative of cos(x) is sin(x), the integral of cos(x) is equal to sin(x) + C.A...
We also get a constant of integration after solving the indefinite integration. Answer and Explanation: {eq}I=\int\sin^{5}2t\cos^{2}2tdt\ 2t=u\ I=\frac{1}{2}\int \sin^4u \cos^2u \sin udu\ =\frac{1}{2}\int(1-\cos^{2}u)^2\cos^2u \sin u du\ \cos......
up出这个对比应该是想表达,两个不定积分形式上虽然有点像,但是实际上结果差异是很大的。 2022-01-22 17:023回复 故事说给枕头听- 第一题分子分母同除cos²x,分子凑微分dtanx,分母sec²x利用三角恒等式化为tan²x+1分母就是2tan²x+1然后套arctan 的积分公式,口算题 2022-01-26 19:21回复 ...
It is often useful to consider a more specific form for IEs, where the functionGfactors in the product of a kernelKand a generally nonlinear functionFasG(y, t, s) = K(t, s)F(y). HereKis matrix valued, and it carries the dependence on time (bothtands), whereasFdepends...
intsinx,x −cosx (2) > intxx3−1,x −lnx2+x+16+3arctan2x+1333+lnx−13 (3) > intexp−x2,x πerfx2 (4) If Maple cannot find a closed form expression for the integral, the function call is retur...
(sinx+cosx)2 dx=∫0π4lnsin(2x)sin2(x+π4) dx=x→π4−x∫0π4sec2xlncos(2x) dx=∫0π4lncos(2x) dtanx=limx→π4lncos(2x)+2∫0π41−cos(2x)cos(2x) dx=−π2+limx→π4ln[1+sin(2x)]=−π2+ln...