What is the antiderivative of sinx? - Socratic
2015年5月29日 · The general antiderivative of sin(x) is -cos(x)+C. With an integral sign, this is written: \\int sin(x)\\ dx=-cos(x)+C.
What is the antiderivative of #sin^2(x)#? - Socratic
2016年7月4日 · How do I evaluate the indefinite integral #intx*sin(x)*tan(x)dx# ? See all questions in Integrals of Trigonometric Functions Impact of this question
What is the antiderivative of sinx^3? - Socratic
2017年6月13日 · If you mean (sinx)^3, please see below. If you mean sin(x^3), I can't help. Here is one possibility int (sinx)^3 dx = int sin^2x sinx dx = int ((1-cos^2 x) sinx dx = int sinx dx + int underbrace(cos^2x)_u underbrace((-sinx) dx)_(du) = -cosx + cos^3x/3 +C As is typical of expressions involving trigonometric functions, there are other ways to express this family of …
What is the antiderivative of (sinx)^2? - Socratic
2016年5月3日 · So, instead of integrating sin^2x, we can integrate 1/2(1-cos2x) and see if that gets us anywhere: int1/2 ...
What is the antiderivative of #sin (lnx)#? - Socratic
2016年1月18日 · In the given context, finding the antiderivative of #sin(ln(x))# is equivalent to solving the indefinite integral #intsin(ln(x))dx# To do so, we will make use of integration by substitution and integration by parts. We will need to do some additional tricks with these beyond the standard basic methods as well. To begin, we let #I = intsin(ln(x))dx#
What is the antiderivative of #(sinx)/(cos^2x)#? - Socratic
2016年6月21日 · A more methodical way, if you are just starting out, would be, starting with the integral #\int \frac{\sin x}{\cos^{2} x} dx #, to make the sub #u = \cos x, du = -\sin x dx# So the integral becomes: #\int \frac{\sin x}{\cos^2 x} (-\frac{1}{\sin x}) du = -\int \frac{1}{u^2} du #
What is the antiderivative of 1/sinx? - Socratic
2016年9月19日 · So the antiderivative is minus the natural logarithm of the denominator. ... #intsinx/sin^2xdx# #intsinx ...
What is the antiderivative of #arcsin(x)#? - Socratic
2016年2月12日 · Let #t = arcsin(x) => sin(t) = x# Then #dx = cos(t)dt# Making the substitution, we have. #int arcsin(x)dx = int tcos(t)dt# #color(white)# Next, we use integration by parts: Let #u = t# and #dv = cos(t)dt# Then #du = dt# and #v = sin(t)# Applying the integration by parts formula #intudv = uv - intvdu# #inttcos(t)dt = tsin(t) - intsin(t)dt#
How do you find the antiderivative of sinx? - Socratic
2018年4月10日 · The antiderivative of sinx is its integral. The integral of #sinx# is a standard results and evaluates to . #intsinxdx=-cosx+"c"#
What is the antiderivative of (sin(x))^2? - Socratic
2016年8月31日 · #cos 2A = cos^2 A - sin^2 A# #= 2 cos^2 A -1# #= 1 - 2 sin^2 A#. last one gives us #sin^2 A = 1/2(1- cos 2A)# So. #int (sin(x))^2 dx# # = 1/2 int 1 - cos 2x dx# # = 1/2 (x -1/2 sin 2x) + C#. which you can flip using the other double-angle formula: #sin 2A = 2 sin A cos A# # = 1/2 (x - sin x cos x) + C#