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In integral calculus, the Weierstrass substitution or tangent halfangle substitution is a method for evaluating integrals, which converts a rational function of trigonometric functions of into an ordinary rational function of by setting = ⁡ (/) No generality is lost by taking these to be rational functions of the sine and cosine The general transformation formula is A) Integral product of two functions = first function × integral of the second function – integral of {differential coefficient of the first function × integral of the second function} ∫ f1(x) f2(x) = f1(x)∫ f2(x)dx − ∫ d dxf1(x) ∫ f2(x)dxdx B) ∫ exf(x) f ′ (x)dx = ∫ exf(x)dx C Here is a summary for this final type of trig substitution √a2b2x2 ⇒ x = a b tanθ, −π 2 < θ < π 2 a 2 b 2 x 2 ⇒ x = a b tan ⁡ θ, − π 2 < θ < π 2 Before proceeding with some more examples let's discuss just how we knew to use the substitutions that

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Tan^2x formula integration-To find some integrals we can use the reduction formulas These formulas enable us to reduce the degree of the integrand and calculate the integrals in a finite number of steps Below are the reduction formulas for integrals involving the most common functions ∫xnemxdx = 1 m xnemx − n m ∫xn−1emxdx ∫ emx xn dx = − emx (n−1)xn−1 Integration of Cos2x The integral of cos2x = (1/2)sin(2x) C, Where C is constant Finding the integral of Cos (2x) f (x) = cos (x) g (x) = 2x f (g (x))= cos (2x) Let u = g (x) =2x then du = g' (x) = 2dx, and d (x) = (1/2)du Plug in u for 2x and (1/2)du for dx

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So tan^2(x)=sec^2(x)1;integrate this equation on both sides w r to x We know that integration of sec^2(x) is tanx c; The integral of $$\int\tan^n(x)dx$$ I found this proof online for the answer and it went like this $$\int\tan^{n2}(x)\tan^2(x)dx$$ $$\int\tan^{n2}(x)(\sec^2(x)1)dx$$ $$\int\tan^{n2}(x)\sec^2(x)dx\int\tan^{n2}(x)dx$$ Then using usubstitution it yielded the reduction formula $$\frac{\tan^{n1}(x)}{n1}\int\tan^{n2}dx$$C being a constant So integration of tan^2(x) would be tanx x c;

 Figure 711 To find the area of the shaded region, we have to use integration by parts For this integral, let's choose u = tan − 1x and dv = dx, thereby making du = 1 x2 1 dx and v = x After applying the integrationbyparts formula (Equation 712) we obtain Area = xtan − 1x1 0 − ∫1 0 x x2 1 dxHow to find the integral of tan(2x)In this tutorial we go through the steps to find the integral of tangent(2x) using the usubstitution integration method1 Integration By Substitution (Change of Variables) We can think of integration by substitution as the counterpart of the chain rule for di erentiation Suppose that g(x) is a di erentiable function and f is continuous on the range of g Integration by substitution is given by the following formulas Inde nite Integral Version Z f(g(x))g0(x)dx= Z

 You can start by writing tan^2(x)=sin^2(x)/cos^2(x) giving inttan^2(x)dx=intsin^2(x)/cos^2(x)dx= using sin^2(x)=1cos^2(x) you get =int(1cos^2(x))/(cos^2(x))dx=int1/cos^2(x)1dx= =int1/cos^2(x)dxint1dx= =tan(x)xcIntegrals (inverse functions) Derivatives; We know that `tan^2(x)=sec^2(x)1` hence we can rewrite our integral in the following manner `inttan^n(x)dx=inttan^(n2)(x)*tan^2(x)dx=` `inttan^(n2)(x)(sec^2(x)1)dx=`

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 Ex 74, 9 sec 2 tan 2 4 Let tan = Diff both sides wrt x sec 2 = = sec 2 Integrating the function sec 2 tan 2 4 Putting value of tan = and = sec 2 = sec 2 tIntroduction to Tan double angle formula let's look at trigonometric formulae also called as the double angle formulae having double angles Derive Double Angle Formulae for Tan 2 Theta \(Tan 2x =\frac{2tan x}{1tan^{2}x} \) let's recall the addition formulaIt may be necessary to repeat this process on the tan k − 2 x tan k − 2 x term If k k is even and j j is odd, then use tan 2 x = sec 2 x − 1 tan 2 x = sec 2 x − 1 to express tan k x tan k x in terms of sec x sec x Use integration by parts to integrate odd powers of sec x sec x

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 Introduction to Integral formulas The list of integral calculus formula is here with all the rules which are needed to solve integration The formula sheet of integration include basic integral formulas, integration by parts and partial fraction, area as a sum and properties of definite integralAt first take a look at indefinite integrationIntegration Reduction Formulas Any positive integer power of sin x can be integrated by using a reduction formula Example Prove that for any integer n 2, Z sin n xdx= 1 n sin C to find the integrals of tan2 x, tan3 x, tan4 x and tan5 x Using the identity tan2 x =sec2 x 1, Z tan n xdx= Z tan n2 xtan2xdx= Z tan n2 x(sec2x 1) dx = Z tan$\int ( \sec^2 x) = \tan x C$ $\int (\csc^2 x) = \cot^2 x C$ $\int ( \sec (x) \tan (x) )=\sec x C$ $\int (\csc (x) \cot (x)) = \csc x C$ $ \int (\tan x) = ln \sec x C$ $ \int (\cot x) = ln \sin x C$ $ \int (\sec x) = ln \sec x \tan x C$ $ \int (\csc x) = ln \csc x – \cot x C$ Integration formulas Related to Inverse Trigonometric Functions

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∫ csc x ( cot x) dx = – csc x C; `int ((3 ln\ 2x)^3)/(x) dx` Let u = 3 ln 2x We can expand out the log term on the right hand side as follows 3 ln 2x = 3 ln 2 ln x Now the first 2 terms on the right are constants (whose derivative equals zero) and the derivative of the natural log of x is `1/x` Then `du = 1/x dx`In trigonometry, tangent halfangle formulas relate the tangent of half of an angle to trigonometric functions of the entire angle The tangent of half an angle is the stereographic projection of the circle onto a line Among these formulas

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∫ sec 2 x dx = tan x C; Chapter 7 Class 12 Integration Formula Sheet by teachoocom Basic Formulae = ^( 1)/( 1) , 1 , = = sin x C = cos x C 2 = tan x c 2 = cot x c = sec x c∫ (1/x) dx = ln x C;

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 In this section we look at integrals that involve trig functions In particular we concentrate integrating products of sines and cosines as well as products of secants and tangents We will also briefly look at how to modify the work for products of these trig functions for some quotients of trig functionsFormula cos ⁡ 2 θ = 1 − tan 2 ⁡ θ 1 tan 2 ⁡ θ A mathematical identity that expresses the expansion of cosine of double angle in terms of tan squared of angle is called the cosine of double angle identity in tangentTo integrate tan^22x, also written as ∫tan 2 2x dx, tan squared 2x, (tan2x)^2, and tan^2(2x), we start by utilising standard trig identities to change the form of the integral Our goal is to have sec 2 2x in the new form because there is a standard integration solution for that in formula booklets that we can use We recall the Pythagorean trig identity, and multiply the angles by 2 throughout

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 u= tan^3 (x) dv=1 du= 3tan 2 (x)sec 2 (x)dx v = x = xtan 3 ∫x3tan 2 (x)sec 2 (x)dx again, ∫x3tan 2 (x)sec 2 (x)dx u = x dv= tan 3 sec 2 du = dx v = (tan ^4 )/¼ This is going no where to what the reduction formula is supposed to look likeReduction formula for In = ∫tannxdx Let u(x) = tann − 2x v ′ (x) = tan2x Integrating by parts In = (tanx − x)(tann − 2x) − ∫(tan(x) − x)(n − 2)(tann − 3x tann − 1x)dx Let the second integral be J, After expanding and simplifying I get, J = (n − 2)In − 2Ln ln 2 2 tan(x a dx x x a x a2 2 2 2 1) ( ) x a ∫ = − − sin ln sin ln cos ln( ) ( )( ) ( ) 2 x ∫ x dx x x= − cos ln sin ln cos ln( ) ( )( ) ( ) 2 x ∫ x dx x x=

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V = cos xTo integrate tan2x, also written as ∫tan2x dx, and tan 2x, we use the u substitution because the integral of tanu is a standard solution in formula books Let u=2x Then, du/dx=2 We transpose for dx to get the above expression Hence, our new integration can be writtin in terms of u and is simpler to solve©05 BE Shapiro Page 3 This document may not be reproduced, posted or published without permission The copyright holder makes no representation about the accuracy, correctness, or

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598 integration techniques If the exponent of cosine is odd, split off one cos(x) and use the identity cos2(x) = 1 sin2(x) to rewrite the remaining even power of cosine in terms of sine Then use the change of variable u = sin(x) If both exponents are even, use the identities sin2(x) = 1 2 1 2 cos(2x) and cos2(x) = 1 2 1 2 cos(2x) to rewrite the integral in terms of powersSo the required answer is tanxxc Where c is any constantSubstitution\\int\frac {e^ {x}} {e^ {x}e^ {x}}dx,\u=e^ {x} integralcalculator \int\tan^ {2} (x)dx en Sign In Sign in with Office365 Sign in with Facebook OR

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∫ sec x (tan x) dx = sec x C;Click here👆to get an answer to your question ️ Integrate the following tanxtan2xtan3x =•use trigonometric identities to integrate sin2 x, cos2 x, and functions ofthe formsin3x cos4x •integrate products of sines and cosines using a mixture of trigonometric identities and integration by substitution •use trigonometric substitutions to evaluate integrals Contents 1 Introduction 2 2

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Current Location > Math Formulas > Calculus > Integrals of Trigonometric Functions Integrals of Trigonometric Functions Don't forget to try our free app Agile Log sec 2 x tan x c csc 2 xcot x c Example 1 Calculate the following integral ∫x 2 sin x 3 dx Solution∫ a x dx = (a x /ln a) C ; If u still do not understand follow these procedures 1) Use tan (ab) to bring the entire integral in terms of tanx alone Note you might have to use tan2x = 2tanx/sec^2x 2)After you have got an expression in terms of tanx integral { (4tan^3x)/ (sec^2x (Sec^2x2tan^2x)) (2tan^3x)/ (sec^2x2tan^2x)}

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Sin ( ) 1 cos(2 )x x ii 12 2 cos ( ) 1 cos(2 )x x If there are no sec(x) factors and the power of tan(x) is even and positive, use sec 1 tan22x x to convert one tan2 x to sec2 x Rules for sec(x) and tan(x) also work for csc(x) andSimilarly to the previous examples, this type of integrals can be simplified by the formula \ If the power of the secant \(n\) is even, then using the identity \(1 {\tan ^2}x \) \(= {\sec ^2}x\) the secant function is expressed as the tangent function The factor \({\sec ^2}x\) is separated and used for transformation of the differential∫ e x dx = e x C;

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As there is no way to immediately integrate tan^2 (x) using well known trigonometric integrals and derivatives, it seems like a good idea would be writing tan^2 (x) as sec^2 (x) 1 Now, we can recognise sec^2 (x) as the derivative of tan (x) (you can prove this using the quotient rule and the identity sin^2 (x) cos^2 (x) = 1), while we getIt does not contain any constant of integration Integral of cos^2x We can't just integrate cos^2(x) as it is, so we want to change it into another form, which we can easily do using trig identities Integral of cos^2(2x) Recall the double angle formula cos(2x) = cos^2(x) – sin^2(x) We also know the trig identity sin^2(x) cos^2(x) = 1,These integral formulas are equally important as differentiation formulas Some other important integration formulas are Classification of Integral Formulas

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4 Chapter 10 Techniques of Integration EXAMPLE 1012 Evaluate Z sin6 xdx Use sin2 x = (1 − cos(2x))/2 to rewrite the function Z sin6 xdx = Z (sin2 x)3 dx = Z (1− cos2x)3 8 dx = 1 8 Z 1−3cos2x3cos2 2x− cos3 2xdx Now we have four integrals to evaluate Z 1dx = x and Z Integral of u^2 is NOT (u^3)/3 c Rather, integral of (u^2)du = (u^3)/3 c In (tan^2)x your 1st mistake is not writing dx Note that dx is NOT always du!!!!!∫ csc 2 x dx = cot x C;

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Solve the integral = ln u C substitute back u=cos x = ln cos x C QED 2 Alternate Form of Result tan x dx = ln cos x C = ln (cos x)1 C = ln sec x C Therefore &=∫(\tan^2x)^2\sec^2x\sec x\tan x\,dx & & \text{Write }\tan^4x=(\tan^2x)^2\\4pt &=∫(\sec^2x−1)^2\sec^2x\sec x\tan x\,dx & & \text{Use }\tan^2x=\sec^2x−1\\4pt &=∫(u^2−1)^2u^2du & & \text{Let }u=\sec x\text{ and }du=\sec x\tan x\,dx\\4ptIf you let u=tanx in integral (tan^2)x you get integral u^2 dx which is not (u^3)/3 c since du= sec^2x dx

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The reduction formulae can be extended to a range of functions The procedure, however, is not the same for every function Eg1 Find a reduction formula to integrate ∫sin n x dx and hence find ∫sin 4 x dx ∫sin n x dx = ∫sin n1 x sin x dx Let u = ∫sin n1 x and dv/dx = sin x So, du/dx = (n1)sin n2 x cos x;SOLUTION We could evaluate this integral using the reduction formula for (Equation 567) together with Example 3 (as in Exercise 33 in Section 56), but a better method is to write and use a halfangle formula tan x tan2x sec2x 1 tan2x y tan3x dx y sec x dx ln sec x tan x C

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