How to solve integral of sec 3x
WebThe formula for the integration of sec 3x can be written as ∫sec 3x dx = (1/3) ln( tan 3x + sec 3x ) + C with C as the constant of integration. How to Find Integration of Sec 3x? We can … WebThe formula for integral (definite) goes like this: $$\int_b^a f(x)dx$$ Where, ∫ represents integral. dx represents the differential of the 'x' variable. fx represents the integrand. point a and b represent limits of integration. Let's solve it considering that we're being asked for integral from 1 to 3, of 3x dx $$\int_3^1 3(x)dx$$ Solving:
How to solve integral of sec 3x
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WebThe formula of cos3x is cos3x = 4 cos^3x - 3 cos x. The derivative of cos3x is -3 sin 3x and the integral of cos3x is (1/3) sin3x + C. The period of cos3x is 2π/3. The most commonly used formula of cos cube x is cos^3x = (1/4) cos3x + (3/4) cosx which is used for simplifying complex integration problems. Web∫sec3(x)dx Recall the identity sec2(x) = tan2(x) + 1. So, substituting, you get ∫sec(x)(tan2(x) + 1)dx = ∫tan(x)tan(x)sec(x)dx + ∫sec(x)dx. The first integral can be solved by u -substitution and integration by parts, while the second, is an identity. ∫tan(x)dsec(x) = tan(x)sec(x) − ∫sec(x)dtan(x) But ∫ sec(x)dtan(x) is the original integral.
WebYou are going to see more and more of this if you continue in math, that is, the creative use of the rules and properties of numbers and processes. In this case, treating the 1 as the result of differentiating some function g(x)=x, made it possible the use of integration by parts to solve the problem. Use whatever works to solve problems. WebDec 21, 2014 · since ∫sec3xdx = I, = secxtanx −I +∫secxdx. by adding I and ∫secxdx = ln secx +tanx +C1. ⇒ 2I = secxtanx + ln secx + tanx + C1. by dividing by 2, ⇒ I = 1 2 secxtanx + 1 2 ln secx + tanx + C1 2. Hence, ∫sec3dx = 1 2secxtanx + 1 2ln secx +tanx + C. I hope that …
Web∫sec 3xdx Hard Solution Verified by Toppr I=∫sec 3xdx Integrating by parts, we have u=secx, dv=sec 2xdx ⇒du=secxtanxdx , ⇒v=tanx =secxtanx−∫secxtan 2xdx =secxtanx−∫secx(sec 2x−1)dx =secxtanx−∫sec 3x−secxdx Since ∫sec 3xdx=I =secxtanx−I+∫secxdx ⇒2I=secxtanx+ln∣secx+tanx∣+c 1 ⇒I= 21secxtanx+ 21ln∣secx+tanx∣+ 2c 1 WebOct 14, 2024 · Let u= sec ( x) 2.Use integration by parts Now doing the first way results in the integrand looking like ∫ u 3 d u = 1 4 sec 4 ( x) + C Which is correct but it's not the answer I'm looking for, so instead we'll do it the second way. ∫ sec 2 ( x) ⋅ sec 2 ( x) tan ( x) d x ∫ ( tan 2 ( x) + 1) sec 2 ( x) tan ( x) d x
WebCan you solve integrals by calculator? Symbolab is the best integral calculator solving indefinite integrals, definite integrals, improper integrals, double integrals, triple integrals, multiple integrals, antiderivatives, and more. Which is an antiderivative? An antiderivative of function f(x) is a function whose derivative is equal to f(x).
WebHow to solve the following indefinite integral $$\int \tan^{3}x \sec^{3/2}x \; dx$$ to get the solution in the form of $$\large\frac{2}{7}\sec^{7/2}x - \frac{2}{3}\sec^{3/2}x +c$$ graphics and multimedia software slideshareWebJun 13, 2024 · Explanation: Suppose that, I = ∫ sec3x tanx dx. ∴ I = ∫ 1 cos3x ⋅ cosx sinx dx = ∫ 1 cos2xsinx dx. = ∫ sinx cos2x ⋅ sin2x dx, = ∫ sinx cos2x(1 − cos2x) dx, Letting cosx = t, so that, −sinxdx = dt, we have, I = ∫ −1 t2(1 − t2) dt = ∫ 1 t2(t2 − 1) dt, = ∫{ t2 −(t2 − 1) t2(t2 −1) dt, = ∫{ t2 t2(t2 − 1) − t2 − 1 t2(t2 −1) }dt, = ∫{ 1 t2 −1 − 1 t2 }dt, chiropractic littleton nhWebHow to integrate sec(3x) The integration of sec 3x is given by (1/3) ln( tan 3x + sec 3x ) + C, where C is the integration constant. It can be calculated using the substitution method. graphics and layout examplesWebTo start with this integral, there is a very powerful trick that allows us to solve it, we will multiply the \sec x by a fraction that contains \sec x + \tan x in the denominator and in the numerator. \displaystyle \int \sec x \cfrac{\sec x + \tan x}{\sec x + \tan x} dx. We will do the multiplication and the obtained result will be the following: chiropractic liveWebtan3xsec3 / 2x = (1 − cos2x)sinxcos − 9 / 2x so that we have ∫tan3xsec3 / 2xdx = ∫(cos − 9 / 2x − cos − 5 / 2x)sinxdx = 2 7cos − 7 / 2x − 2 3cos − 3 / 2x + C Share Cite Follow answered Jul 2, 2015 at 18:11 Mark Viola 173k 12 138 239 Add a comment You must log in to answer this question. Not the answer you're looking for? graphics and multimedia software meaningWebCos 3x is an important formula in trigonometry given by cos 3x = 4 cos 3 x - 3 cos x. Now, we will use this formula to calculate the integration of cos 3x. But before this, we will evaluate the integral of cos cube x, that is, determine the value of ∫cos 3 x dx. To determine this integral, we will use trigonometric formulas and identities. chiropractic lithographWebAug 19, 2016 · I=intcsc^3xdx We will use integration by parts. First, rewrite the integral as: I=intcsc^2xcscxdx Since integration by parts takes the form intudv=uv-intvdu, let: Applying integration by parts: I=-cotxcscx-intcot^2xcscxdx Through the Pythagorean identity, write cot^2x as csc^2x-1. I=-cotxcscx-int (csc^2x-1) (cscx)dx graphics and media panel