Integrate sin^2x cos^4x dx

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Integrate sin^2x cos^4x dx

\(\displaystyle \int sin^2xcos^4xdx\) \(\displaystyle \int(1-cos^2x)(cos^4x)\) \(\displaystyle \int(cos^4x-cos^6x)dx\) \(\displaystyle \cos^2 x = \frac {1 + \cos 2x}2\) \(\displaystyle \frac{1}{2}\int cos^4xdx-\frac{1}{2}\int cos^6xdx\) \(\displaystyle \frac{1}{8}\int (1+cos2x)^2-\frac{1}{16}\int(1+cos2x)^3\) I'm stuck here because it seems as if l am chasing my tail. Re: Integration of trig functions \(\displaystyle \int sin^{2}(x)cos^{4}(x)dx\) \(\displaystyle =\frac{1}{8}\int (1-cos(2x))(1+cos(2x))^{2}dx\) \(\displaystyle =\frac{1}{8}\int (1-cos^{2}(2x))(1+cos(2x))dx\) \(\displaystyle =\frac{1}{8}\int sin^{2}(2x)dx+\frac{1}{8}\int sin^{2}(2x)cos(2x)dx\) \(\displaystyle =\frac{1}{16}\int (1cos(4x))dx+\frac{1}{48}sin^{3}(2x)\) \(\displaystyle \boxed{=\frac{1}{16}x-\frac{1}{64}sin(4x)+\frac{1}{48}sin^{3}(2x)+C}\) Re: Integration of trig functions \(\displaystyle \int sin^{2}(x)cos^{4}(x)dx\) \(\displaystyle =\frac{1}{8}\int (1-cos(2x))(1+cos(2x))^{2}dx\) \(\displaystyle =\frac{1}{8}\int (1-cos^{2}(2x))(1+cos(2x))dx\) \(\displaystyle =\frac{1}{8}\int sin^{2}(2x)dx+\frac{1}{8}\int sin^{2}(2x)cos(2x)dx\) \(\displaystyle =\frac{1}{16}\int (1-cos(4x))dx+\frac{1}{48}sin^{3}(2x)\) \(\displaystyle \boxed{=\frac{1}{16}x-\frac{1}{64}sin(4x)+\frac{1}{48}sin^{3}(2x)+C}\) Galactus how did you get : \(\displaystyle \frac{1}{48}sin^{3}(2x)+C}\) Setting $u = \sin(x), \; dv = \cos^4(x)\sin(x) \, dx$, we get $$\int \sin^2(x) \cos^4(x) \, dx = -\sin(x)\frac{\cos^5(x)}{5} + \frac{1}{5}\int \cos^6(x) \, dx$$ which certainly doesn't look any better, but we can take two cosines from the six and convert them into sines to get $$\int (1-\sin^2(x)) \cos^4(x) \, dx = \int \cos^4(x) \,dx - \int\sin^2(x) \cos^4(x) \, dx$$ but that's the same as our original integral, so we can move it over to the LHS to get $$\frac{6}{5} \int \sin^2(x) \cos^4(x) \, dx = -\sin(x)\frac{\cos^5(x)}{5} + \frac{1}{5}\int \cos^4(x) \, dx$$ so that we now have a reduced problem. In fact, you could even do a similar trick to evaluate the remaining integral. sin2(x)cos4(x)dxRemember that the half angle identities are:sin(/2) = (1-cos())/2 so squaring both sides gives:sin2(/2) = (1 - cos())/2 and letting x = /2 meaning that = 2x we have:sin2(x) = (1 - cos(2x))/2 = 1/2(1 - cos(2x))The same holds true for the cosine half angle formula so:cos2(x) = 1/2(1 + cos(2x))Substituting these into the integral, we get:[1/2(1 - cos(2x))][1/2(1 + cos(2x))][1/2(1 + cos(2x))]dx1/8[1 - cos(2x)][1 + cos(2x)][1 + cos(2x)]dx1/8[1 - cos2(2x)][1 + cos(2x)]dx1/8[1 - cos2(2x) + cos(2x) - cos3(2x)]dx1/81dx - 1/8cos2(2x)dx + 1/8cos(2x)dx - 1/8cos3(2x)dxLet's break this into pieces so we can keep track of it better.Let this become A - B + C - D whereA = 1/81dxB = 1/8cos2(2x)dxC = 1/8cos(2x)dxD = 1/8cos3(2x)dxA is trivial to resolve. A = 1/8xC is pretty easy: C = 1/8[1/2sin(2x)] = 1/16sin(2x)For B, let's again use the half angle equation we used beforeB = 1/8cos2(2x)dxB = 1/81/2(1 + cos(4x))dxB = 1/16[1dx + cos(4x)dx]B = 1/16[x + 1/4sin(4x)]B = 1/16x + 1/64sin(4x)For D, we can use the Pythagorean Identity (sin2(x) + cos2(x) = 1)D = 1/8cos3(2x)dxD = 1/8cos2(2x)cos(2x)dxD = 1/8(1 - sin2(2x))cos(2x)dxD = 1/8cos(2x)dx - 1/8sin2(2x)cos(2x)dxD = 1/16sin(2x) - 1/8sin2(2x)cos(2x)dxTo do the right side of this, use a "u" substitutionLet u = sin(2x) then u' = 2cos(2x)so 1/8sin2(2x)cos(2x)dx = (1/8)1/2sin2(2x)2cos(2x)dx = 1/16u2du = 1/16(1/3u3) = 1/48sin3(2x)so D = 1/16sin(2x) - 1/48sin3(2x)Combining:A - B + C - D1/8x - (1/16x + 1/64sin(4x)) + 1/16sin(2x) - (1/16sin(2x) - 1/48sin3(2x))1/8x - 1/16x - 1/64sin(4x) + 1/16sin(2x) - 1/16sin(2x) + 1/48sin3(2x)1/16x - 1/64sin(4x) + 1/48sin3(2x) + CBy the way, there can be lots of versions of this that are correct. The easiest way to find out if an answer is correct is to plug in integral bounds and see if you get the same numerical answers. 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}{cos^{4}x}dx en Feedback

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