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How to integrate sine squared

Integration of sin^2x: $\displaystyle\int\sin^2xdx=\dfrac{1}{2}x-\dfrac{1}{4}\sin 2x+C$ or $\displaystyle\int\sin^2xdx=\dfrac{1}{2}x-\dfrac{1}{2}\sin x\cos x+C$ We introduce 2 methods to integrate $\sin^2x$ Proof 1: By Half-angle Formula Strategy: Convert $\sin^2x$ to $\cos 2x$ by the following half-angle formula: $\sin^2x=\dfrac{1-\cos 2x}{2}$ Proof: $\displaystyle\int \sin^2xdx\\ =\displaystyle\int\left(\dfrac{1-\cos 2x}{2}\right)dx\\ =\dfrac{1}{2}x-\dfrac{1}{2}\displaystyle\int\cos 2xdx\\ =\dfrac{1}{2}x-\dfrac{1}{2}\cdot\dfrac{\sin 2x}{2}+C$ $=\dfrac{1}{2}x-\dfrac{1}{4}\sin 2x+C$ Note: Integration of $\cos ax$ was used in the above proof: $\displaystyle\int \cos axdx=\dfrac{1}{a}\sin ax+C$ Proof 2: Integration by Parts Strategy: Integration by parts Trigonometric identity: $\sin^2x+\cos^2x=1$ Proof: $\displaystyle\int \sin^2xdx\\ =\displaystyle\int(\sin x)(-\cos x)'dx\\ =(\sin x)(-\cos x)-\displaystyle\int(\cos x)(-\cos x)dx\\ =-\sin x\cos x+\displaystyle\int(\cos^2x)dx\\ =-\sin x\cos x+\displaystyle\int(1-\sin^2x)dx\\ =-\sin x\cos x+x-\displaystyle\int\sin^2xdx+C$ Hence $2\displaystyle\int\sin^2xdx=-\sin x\cos x+x+C$ $\displaystyle\int\sin^2xdx=\dfrac{1} {2}x-\dfrac{1}{2}\sin x\cos x+C$ Relation between Two Results The two integration results of $\sin^2x$ are equivalent: $\displaystyle\int\sin^2xdx=\dfrac{1}{2}x-\dfrac{1}{4}\sin 2x+C$ $\displaystyle\int\sin^2xdx=\dfrac{1}{2}x-\dfrac{1}{2}\sin x\cos x+C$ due to the following double-angle formula: $\sin 2x=2\sin x\cos x$ Notes on Integration Half-angle formulas: $\sin^2x=\dfrac{1-\cos 2x}{2}$ $\cos^2x=\dfrac{1+\cos 2x}{2}$ are useful for the following integration: $\displaystyle\int\sin^2xdx$,$\displaystyle\int\sin^3xdx$,$\displaystyle\int\sin^4xdx$,$\displaystyle\int\sin^2x\cos xdx$,$\displaystyle\int\sin x\cos^2 xdx$ With half-angle formulas, we can turn integral functions into polynomial of lower degrees on $\sin x$ or $\cos x$. Note that integration of $\tan^2x$ and $\tan^3x$ require other methods than half-angle formulas. What is the integral of tan^3x? As soon as you see a question asking you to integrate the square of sin, cos or tan, your first approach should be to use trigonometric identities and double angle formulas. For sin2(X), we will use the cos double angle formula: cos(2X) = 1 - 2sin2(X) The above formula can be rearranged to make sin2(X) the subject: sin2(X) = 1/2(1 - cos(2X)) You can now rewrite the integration: sin2(X)dX = 1/2(1 - cos(2X))dX Because 1/2 is a constant, we can remove it from the integration to make the calculation simpler. We are now integrating: 1/2 x (1 - cos(2X)) dX = 1/2 x (X - 1/2sin(2X)) + C It is very important that as this is not a definite integral, we must add the constant C at the end of the integration. Simplifying the above equation gives us a final answer: sin2(X) dX = 1/2X - 1/4sin(2X) + C In this tutorial we shall derive the integral of sine squared x. The integration is of the form \[I = \int {{{\sin }^2}xdx} \] This integral cannot be evaluated by the direct formula of integration, so using the trigonometric identity of half angle $${\sin ^2}x = \frac{{1 ? \cos 2x}}{2}$$, we have \[\begin{gathered} I = \int {\left( {\frac{{1 ? \cos 2x}}{2}} \right)dx} \\ \Rightarrow I = \frac{1}{2}\int {\left( {1 ? \cos 2x} \right)dx} \\ \Rightarrow I = \frac{1}{2}\int {1dx ? \frac{1}{2}\int {\cos 2xdx} } \\ \end{gathered} \] Using the integral formula $$\int {\cos kxdx = \frac{{\sin kx}}{k} + c} $$, we have \[\begin{gathered} \int {{{\sin }^2}x} dx = \frac{1}{2}x ? \frac{1}{2}\frac{{\sin 2x}}{2} + c \\ \Rightarrow \int {{{\sin }^2}x} dx = \frac{1}{2}x ? \frac{1}{4}\sin 2x + c \\ \end{gathered} \] From Calculus This article is about a particular function from a subset of the real numbers to the real numbers. Information about the function, including its domain, range, and key data relating to graphing, differentiation, and integration, is presented in the article.View a complete list of particular functions on this wiki This function, denoted , is defined as the composite of the square function and the sine function. Explicitly, it is the map: For brevity, we write as . Key data Item Value Default domain all real numbers, i.e., all of range , i.e., absolute maximum value: 1, absolute minimum value: 0 period , i.e., local maximum value and points of attainment All local maximum values are equal to 1, and are attained at odd integer multiples of . local minimum value and points of attainment All local minimum values are equal to 0, and are attained at integer multiples of . points of inflection (both coordinates) odd multiples of , with value 1/2 at each point. derivative , i.e., double-angle sine function. second derivative derivative times an expression that is or of , depending on the remainder of mod antiderivative mean value over a period 1/2 expression as a sinusoidal function plus a constant function important symmetries even function (follows from composite of even function with odd function is even, the square function being even, and the sine function being odd)more generally, miror symmetry about any vertical line of the form , an integer.Also, half turn symmetry about all points of the form . interval description based on increase/decrease and concave up/down For each integer , the interval from to is subdivided into four pieces:: increasing and concave up: increasing and concave down: decreasing and concave down, : decreasing and concave up power series and Taylor series The power series about 0 (which is hence also the Taylor series) is It is a globally convergent power series. Identities We have the following important identities involving : , relating it to the cosine-squared function. , or equivalently, . Graph Here is the graph on the interval , drawn to scale: Here is a close-up view of the graph between and . The dashed horizontal line indicates the mean value of : The red dotted points indicate the points of inflection and the black dotted points indicate local extreme values. Here is a picture showing the function (blue) and the cosine-squared function (purple) with the dashed line being . The picture illustrates that : Differentiation First derivative WHAT WE USE: chain rule for differentiation, differentiation rule for power functions, sine function#First derivative, double angle cosine formula We have: We can do this two ways. Using the chain rule for differentiation, we have: By the double angle sine formula, this is the same as . Alternatively, using the double angle cosine formula, we rewrite: Differentating, we get: Second derivative Differentiating the derivative again, we get: Graph of function with derivative Fill this in later Points and intervals of interest Critical points Consider . As computed earlier, we have: This equals zero precisely at the points where , so . In other words, the critical points occur at the integer multiples of . Intervals of increase and decrease The function is positive for , with and negative for , with . Dividing by 2, we get: is increasing on intervals of the form , . is decreasing on intervals of the form , . Local extreme values From the information on the intervals of increase and decrease, we conclude that: Intervals of concave up and concave down The second derivative is the function . This is positive for and negative for , where . We thus get: is concave up on intervals of the form , with . is concave down on intervals of the form , with . Points of inflection From the determination of intervals where is concave up and concave down, we discover that the points of inflection are the points with -coordinate an odd multiple of . The function value at all these points is . At points with , the function is transitioning from concave up (on the left) to concave down (on the right). At points with , the function is transitioning from concave down (on the left) to concave up (on the right). Integration First antiderivative WHAT WE USE: double angle cosine formula, recursive version of integration by parts, integration of linear transform of function Using the double angle cosine formula We can now do the integration: To integrate , we use the method of integration of linear transform of function to get . Plugging that in, we get: Using integration by parts We rewrite and use integration by parts in its recursive version: We now rewrite and obtain: Setting to be a choice of antiderivative so that the above holds without any freely floating constants, we get: Rearranging, we get: This gives: So the general antiderivative is: Using the double angle sine formula , we can verify that this matches with the preceding answer. For a given continuous function on a connected set, antiderivatives obtained by different methods must differ by a constant. In some cases, the antiderivatives may be exactly equal, but this is not necessary in general.See zero derivative implies locally constant Graph of function with antiderivative In the picture below, we depict (blue) and the function (purple). This is the unique antiderivative that takes the value 0 at 0. The other antiderivatives can be obtained by vertically shifting the purple graph: The black dots correspond to local extreme values for , and the red dots correspond to points of inflection for the antiderivative. Each black dot is in the same vertical line as a red dot, as we should expect, because points of inflection for the antiderivative correspond to local extreme values for the original function. Further: The antiderivative is increasing everywhere because is everywhere nonnegative, and is zero only at isolated points. The antiderivative is concave up on those intervals where is increasing, i.e., intervals of the form as varies over the integers. The antiderivative is concave down on those intervals where is decreasing, i.e., intervals of the form as varies over the integers. Definite integrals The part in the antiderivative signifies that the linear part of the antiderivative of has slope , and this is related to the fact that has a mean value of on any interval of length equal to the period. It is in fact clear that the function is a sinusoidal function about . Thus, we have: where is an integer. The mean value of over an interval of length equal to a multiple of the period is . Thus, for very large intervals, the mean value of is very close to 1/2, even though it need not be exactly 1/2. Specifically: Transformed versions Based on the integration of , we can also integrate the square of any sinusoidal function using the integration of linear transform of function: We thus see that the mean value of this function is also over any interval of length a multiple of the period . Also, over a sufficiently long interval, the mean value is close to 1/2: Higher antiderivatives It is possible to antidifferentiate more than once. The antiderivative is the sum of a polynomial of degree and a trigonometric function with a period of . Power series and Taylor series Computation of power series We can use the identity: along with the power series for the cosine function, to find the power series for . The power series for the cosine function converges to the function everywhere, and is: The power series for is: The power series for is: Dividing by 2, we get the power series for : Here's another formulation with the first few terms written more explicitly: Taylor polynomials as approximations Note that since is an even function, all its Taylor polynomials are also even polynomials. In the figure below, we consider the graphs of and its second, fourth, and sixth Taylor approximations. Second Taylor polynomial , which equals the third Taylor polynomial, , is . Fourth Taylor polynomial , which equals the fifth Taylor polynomial, , is . Sixth Taylor polynomial, , which equals the seventh Taylor polynomial , is . Limit computations Order of zero We get the following limit from the power series: Thus, the order of the zero of at zero is 2 and the residue is 1. This limit can be computed in many ways: Name of method for computing the limit Details Simple manipulation, using Using the L'Hopital rule Using the power series We have , so we get. Taking the limit as gives 1. Higher order limits We have the limit: This limit can be computed in many ways: Name of method for computing the limit Details Using and We have .The first limit is and the second limit is 2 from the given data. We thus get . Using the L'Hopital rule . Using the power series We have , so , so the limit as is 1/3.

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