Int(sin^(2)x)/(1 cos x)dx is equal to

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Int(sin^(2)x)/(1 cos x)dx is equal to

The integral of sin 2x and the integral of sin2x have different values. To find the integral of sin2x, we use the cos 2x formula and the substitution method whereas we use just the substitution method to find the integral of sin 2x. Let us identify the difference between the integral of sin 2x and the integral of sin2x by finding their values using appropriate methods and also we will solve some problems related to these integrals. What is the Integral of Sin 2x dx? The integral of sin 2x is denoted by sin 2x dx and its value is -(cos 2x) / 2 + C, where 'C' is the integration constant. For proving this, we use the integration by substitution method. For this, we assume that 2x = u. Then 2 dx = du (or) dx = du/2. Substituting these values in the integral sin 2x dx, sin 2x dx = sin u (du/2) = (1/2) sin u du We know that the integral of sin x is -cos x + C. So, = (1/2) (-cos u) + C Substituting u = 2x back here, sin 2x dx = -(cos 2x) / 2 + C This is the integral of sin 2x formula. Definite Integral of Sin 2x A definite integral is an indefinite integral with some lower and upper bounds. By the fundamental theorem of Calculus, to evaluate a definite integral, we substitute the upper bound and the lower bound in the value of the indefinite integral and then subtract them in the same order. While evaluating a definite integral, we can ignore the integration constant. Let us calculate some definite integrals of integral sin 2x dx here. Integral of Sin 2x From 0 to pi/2 \(_0^{\pi/2}\) sin 2x dx = (-1/2) cos (2x) \(\left. \right|_0^{\pi/2}\) = (-1/2) [cos 2(/2) - cos 2(0)] = (-1/2) (-1 - 1) = (-1/2) (-2) = 1 Therefore, the integral of sin 2x from 0 to pi/2 is 1. Integral of Sin 2x From 0 to pi \(_0^{\pi}\) sin 2x dx = (-1/2) cos (2x) \(\left. \right|_0^{\pi}\) = (-1/2) [cos 2() - cos 2(0)] = (-1/2) (1 - 1) = 0 Therefore, the integral of sin 2x from 0 to pi is 0. What is the Integral of Sin^2x dx? The integral of sin2x is denoted by sin2x dx and its value is (x/2) - (sin 2x)/4 + C. We can prove this in the following two methods. By using the cos 2x formula By using the integration by parts Method 1: Integral of Sin^2x Using Double Angle Formula of Cos To find the integral of sin2x, we use the double angle formula of cos. One of the cos 2x formulas is cos 2x = 1 - 2 sin2x. By solving this for sin2x, we get sin2x = (1 - cos 2x) / 2. We use this to find sin2x dx. Then we get sin2x dx = (1 - cos 2x) / 2 dx = (1/2) (1 - cos 2x) dx = (1/2) 1 dx - (1/2) cos 2x dx We know that cos 2x dx = (sin 2x)/2 + C. So sin2x dx = (1/2) x - (1/2) (sin 2x)/2 + C (or) sin2x dx = x/2 - (sin 2x)/4 + C This is the integral of sin^2 x formula. Let us prove the same formula in another method. Method 2: Integral of Sin^2x Using Integration by Parts We know that we can write sin2x as sin x ? sin x. To find the integral of a product, we can use the integration by parts. sin2x dx = sin x ? sin x dx = u dv Here, u = sin x and dv = sin x dx. Then du = cos x dx and v = -cos x. By integration by parts formula, u dv = uv - v du sin x ? sin x dx = (sin x) (-cos x) - (-cos x)(cos x) dx sin2x dx = (-1/2) (2 sin x cos x) + cos2x dx By the double angle formula of sin, 2 sin x cos x = sin 2x and by a trigonometric identity, cos2x = 1 - sin2x. So sin2x dx = (-1/2) sin 2x + (1 - sin2x) dx sin2x dx = (-1/2) sin 2x + 1 dx - sin2x dx sin2x dx + sin2x dx = (-1/2) sin 2x + x + C 2 sin2x dx = x - (1/2) sin 2x + C sin2x dx = x/2 - (sin 2x)/4 + C (where C = C/2) Hence proved. Definite Integral of Sin^2x To evaluate the definite integral of sin2x, we just substitute the upper and lower bounds in the value of the integral of sin2x and subtract the resultant values. Let us evaluate some definite integrals of integral sin2x dx here. Integral of Sin^2x From 0 to 2pi \(_0^{2\pi}\) sin2x dx = [x/2 - (sin 2x)/4] \(\left. \right|_0^{2\pi}\) = [2/2 - (sin 4)/4] - [0 - (sin 0)/4] = - 0/4 = Therefore, the integral of sin2x from 0 to 2 is . Integral of Sin^2x From 0 to pi \(_0^{\pi}\) sin2x dx = [x/2 - (sin 2x)/4] \(\left. \right|_0^{\pi}\) = [/2 - (sin 2)/4] - [0 - (sin 0)/4] = /2 - 0/4 = /2 Therefore, the integral of sin2x from 0 to is /2. Important Notes Related to Integral of Sin 2x and Integral of Sin2x: sin 2x dx = -(cos 2x)/2 + C sin2x dx = x/2 - (sin 2x)/4 + C Topics Related to Integral of Sin2x and Integral of Sin 2x: Example 1: Evaluate the integral sin 2x ecos 2x dx. Solution: We are going to solve this using the integration by substitution. Let cos 2x = u. Then -2 sin 2x dx = du (or) sin 2x dx = (-1/2) du. Then the given integral becomes, eu (-1/2) du = (-1/2) eu + C. Let us substitute the value of u = cos 2x back here. Then we get sin 2x ecos 2x dx = (-1/2) ecos 2x + C. Answer: The integral of sin 2x e^cos 2x dx is (-1/2) ecos 2x + C. Example 2: Evaluate the integral sin2x cos2x dx. Solution: By double angle formula of sin, 2 sin A cos A = sin 2A. Using this, sin2x cos2x dx = (1/4) (2 sin x cos x)2 dx = (1/4) sin2(2x) dx We have sin2x = (1 - cos 2x)/2. From this, sin22x = (1 - cos 4x)/2. So the above integral becomes = (1/4) (1 - cos 4x)/2 dx = (1/8) 1 dx - (1/8) cos 4x dx To solve the second integral, we assume 4x = u, from which 4dx = du (or) dx = du/4. Then we get = (1/8) (x) - (1/8) cos u (du/4) = x/8 - (1/32) cos u du = x/8 - (sin 4x) / 32 + C Answer: The integral of integral sin^2x cos^2x dx is x/8 - (sin 4x) / 32 + C. Example 3: Evaluate the integral sin2x cos3x dx. Solution: The given integral can be written as sin2x cos2x (cos x dx) We know that cos2x = 1 - sin2x. Then the above integral becomes sin2x (1 - sin2x) (cos x dx) = (sin2x - sin4x) (cos x dx) Assume that sin x = u. Then cos x dx = du. = (u2 - u4) du = u3/3 - u5/5 + C Substituting u = sin x back, = sin3x/3 - sin5x/5 + C Answer: The integral of sin^2x cos^3x dx is sin3x/3 - sin5x/5 + C. View Answer > go to slidego to slidego to slide Great learning in high school using simple cues Indulging in rote learning, you are likely to forget concepts. With Cuemath, you will learn visually and be surprised by the outcomes. Book a Free Trial Class FAQs on Integral of Sin 2x and Sin^2x The integral of sin 2x dx is written as sin 2x dx and sin 2x dx = -(cos 2x)/2 + C, where C is the integration constant. What is the Integral of Sin^2x dx? The integral of sin^2x dx is written as sin2x dx and sin2dx = x/2 - (sin 2x)/4 + C. Here, C is the integration constant. How to Find the Definite Integral of Sin 2x from 0 to Pi/4? We know that sin 2x dx = -(cos 2x)/2. Substituting the limits 0 and /4, we get (-1/2) cos 2(/4) - (-1/2) cos 2(0) = (1/2) 0 + (1/2) (1) = 1/2. What is the Integral of Sin^3x dx? sin3x dx = sin2x sin x dx = (1 - cos2x) sin x dx. Let us substitute cos x = u. Then -sin x dx = du. Then the above integral becomes, (1 - u2) (- du) = -u + u3/3 + C. Substituting u = sin x back here, sin3x dx = -cos x + cos3x/3 + C. What is the Integral of Sin 3x dx? To find the sin 3x dx, let that 3x = u. Then 3 dx = du. From this, dx = du/3. Then the above integral becomes, sin u (1/3) du = (1/3) (-cos u) + C = (-1/3) cos (3x) + C. How to Find the Definite Integral of sin^2x from 0 to Pi/4? We know that sin2dx = x/2 - (sin 2x)/4 + C. Substituting the limits here, we get [/8 - (sin /2)/4] - [0 - (sin 0)/4] = /8 - 1/4. Is the Integral sin 2x dx Same as Integral sin^2x dx? No, the values of these two integrals are NOT same. We have sin2dx = x/2 - (sin 2x)/4 + C sin 2x dx = (-cos 2x)/2 + C Something went wrong. Wait a moment and try again. This website uses cookies to ensure you get the best experience. By using this website, you agree to our Cookie Policy. Learn more \bold{\mathrm{Basic}} \bold{\alpha\beta\gamma} \bold{\mathrm{AB\Gamma}} \bold{\sin\cos} \bold{\ge\div\rightarrow} \bold{\overline{x}\space\mathbb{C}\forall} \bold{\sum\space\int\space\product} \bold{\begin{pmatrix}\square&\square\\\square&\square\end{pmatrix}} \bold{H_{2}O} \square^{2} x^{\square} \sqrt{\square} throot[\msquare]{\square} \frac{\msquare}{\msquare} \log_{\msquare} \pi \theta \infty \int \frac{d}{dx} \ge \le \cdot \div x^{\circ} (\square) |\square| (f\:\circ\:g) f(x) \ln e^{\square} \left(\square\right)^{'} \frac{\partial}{\partial x} \int_{\msquare}^{\msquare} \lim \sum \sin \cos \tan \cot \csc \sec \alpha \beta \gamma \delta \zeta \eta \theta \iota \kappa \lambda \mu u \xi \pi \rho \sigma \tau \upsilon \phi \chi \psi \omega A B \Gamma \Delta E Z H \Theta K \Lambda M N \Xi \Pi P \Sigma T \Upsilon \Phi X \Psi \Omega \sin \cos \tan \cot \sec \csc \sinh \cosh \tanh \coth \sech \arcsin \arccos \arctan \arccot \arcsec \arccsc \arcsinh \arccosh \arctanh \arccoth \arcsech \begin{cases}\square\\\square\end{cases} \begin{cases}\square\\\square\\\square\end{cases} = e \div \cdot \times < > \le \ge (\square) [\square] \:\longdivision{} \times \twostack{}{} + \twostack{} {} - \twostack{}{} \square! x^{\circ} \rightarrow \lfloor\square\rfloor \lceil\square\rceil \overline{\square} \vec{\square} \in \forall otin \exist \mathbb{R} \mathbb{C} \mathbb{N} \mathbb{Z} \emptyset \vee \wedge eg \oplus \cap \cup \square^{c} \subset \subsete \superset \supersete \int \int\int \int\int\int \int_{\square}^{\square} \int_{\square}^{\square}\int_{\square}^{\square} \int_{\square}^{\square}\int_{\square}^{\square}\int_{\square}^{\square} \sum \prod \lim \lim _{x\to \infty } \lim _{x\to 0+} \lim _{x\to 0-} \frac{d}{dx} \frac{d^2}{dx^2} \left(\square\right)^{'} \left(\square\right)^{''} \frac{\partial}{\partial x} (2\times2) (2\times3) (3\times3) (3\times2) (4\times2) (4\times3) (4\times4) (3\times4) (2\times4) (5\times5) (1\times2) (1\times3) (1\times4) (1\times5) (1\times6) (2\times1) (3\times1) (4\times1) (5\times1) (6\times1) (7\times1) \mathrm{Radians} \mathrm{Degrees} \square! ( ) % \mathrm{clear} \arcsin \sin \sqrt{\square} 7 8 9 \div \arccos \cos \ln 4 5 6 \times \arctan \tan \log 1 2 3 - \pi e x^{\square} 0 . \bold{=} + \mathrm{simplify} \mathrm{solve\:for} \mathrm{inverse} \mathrm{tangent} \mathrm{line} See All area asymptotes critical points derivative domain eigenvalues eigenvectors expand extreme points factor implicit derivative inflection points intercepts inverse laplace inverse laplace partial fractions range slope simplify solve for tangent taylor vertex geometric test alternating test telescoping test pseries test root test Related ? Graph ? Number Line ? Similar ? Examples ? Our online expert tutors can answer this problem Get step-by-step solutions from expert tutors as fast as 15-30 minutes. Your first 5 questions are on us! In partnership with You are being redirected to Course Hero I want to submit the same problem to Course Hero Examples step-by-step \int\frac{sin^{2} (x)}{1+cosx}dx en Feedback

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