Integralx^2 (x+1)^3 dx=. Question from @Putrijohanti - Sekolah Dasar - Matematika. sebuah akuarium mempunyai volume 240 liter .jika akuarium kosong tersebut di aliri air dengan debit 30 liter/menit,waktu yg di perlukan untuk mengisi akuarium sampai penuh adalah.. a.3menit b.6 menit c.8 menit d.16 menit
$\begingroup$What's the integration of $$\int \sin^5 x \cos^2 x\,dx?$$ Julien44k3 gold badges83 silver badges163 bronze badges asked Feb 3, 2013 at 1949 $\endgroup$ 2 $\begingroup$ Hint Write $$ \sin^5x\cos^2x=\sin^2x^2\cos^2x\sinx. $$ Now use $\cos^2x+\sin^2x=1$ and do the appropriate change of variable. This is the general method to integrate functions of the type $$ \cos^nx\sin^mx $$ when one of the integers $n,m$ is odd. answered Feb 3, 2013 at 1954 JulienJulien44k3 gold badges83 silver badges163 bronze badges $\endgroup$ $\begingroup$ $$ \int \sin^5 x \cos^2x dx $$ $$= \int\sin^2x^2 \cos^2x \sinx dx$$ $$=-\int1 - \cos^2x^2 cos^2x -sinx dx $$ Let $u = \cosx$ $\implies du = -\sinx dx$ $$= -\int1 - u^2² u² du$$ $$= -\int1 - 2u^2 + u^4 u^2 du $$ $$= -\intu^2 - 2u^4+ u^6 du$$ $$= -\left\frac{u^3}{3} - \frac{2u^5}{5} + \frac{u^7}{7}\right + C$$ $$= -u^3\left\frac{1}{3} - \frac{2u^2}{5} +\frac{ u^4}{7}\right + C $$ $$= -\cos^3x \left\frac{1}{3} - \frac{2\cos^2x}{5} + \frac{\cos^4x}{7}\right + C $$ $$= -\cos^3x\frac{15\cos^4x - 42\cos^2x + 35}{105} + C $$ answered Oct 21, 2015 at 1432 $\endgroup$ 1 $\begingroup$ Using trig identities, you can show that $$\sin ^5x \cos ^2x=\frac{5 \sin x}{64}+\frac{1}{64} \sin 3 x-\frac{3}{64} \sin 5 x+\frac{1}{64} \sin 7 x$$ To do this, first use the "Power-reduction formulas" to reduce to get $$\sin^5x=\frac{10 \sin x - 5 \sin 3 x+ \sin 5 x}{16}$$ $$\cos^2x=\frac{1 + \cos 2 x}{2}$$ And then use $$\cos 2 x \sin nx = {{\sinn+2x - \sinn-2x} \over 2}$$ answered Feb 3, 2013 at 2000 gold badges81 silver badges139 bronze badges $\endgroup$ 5 You must log in to answer this question. Not the answer you're looking for? Browse other questions tagged .
Konsepdasar dari metode ini adalah dengan mengubah integral yang kompleks menjadi bentuk yang lebih sederhana. Bentuk umum integral substitusi adalah sebagai berikut. Contoh # 1: Tentukanlah nilai dari : l a t e x ∫ 2 x ( x 2 + 3) 4 d x. Penyelesaian: Langkah pertama kita membuat pemisalan terhadap fungsi yang di integralkan.
\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} \nthroot[\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 fx \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 \nu \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} = \ne \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 \notin \exist \mathbb{R} \mathbb{C} \mathbb{N} \mathbb{Z} \emptyset \vee \wedge \neg \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{Radianas} \mathrm{Graus} \square! % \mathrm{limpar} \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{=} + Inscreva-se para verificar sua resposta Fazer upgrade Faça login para salvar notas Iniciar sessão Mostrar passos Reta numérica Exemplos \int \int \frac{1}{x}dxdx \int_{0}^{1}\int_{0}^{1}\frac{x^2}{1+y^2}dydx \int \int x^2 \int_{0}^{1}\int_{0}^{1}xy\dydx Mostrar mais Descrição Resolver integrais duplas passo a passo double-integrals-calculator \int\sin^{5}\leftx\rightdx pt Postagens de blog relacionadas ao Symbolab High School Math Solutions – Polynomial Long Division Calculator Polynomial long division is very similar to numerical long division where you first divide the large part of the... Read More Digite um problema Salve no caderno! Iniciar sessão
Integral(cos 4 2x dx) Integral (1/2(1-cos 4x)) 2 Integral ([1/4(1-2cos 4x- cos 2 4x)][1/4(1-2cos 4x - cos 2 4x)]dx) 1/64 (24x + 8 sin 4x + 8 sin x) + C . pka Elite Member. Joined Jan 29, 2005 Messages 11,693. Feb 22, 2012 #2 Look at this website. Be sure to click the show steps button. S. soroban

Calculus Examples Popular Problems Calculus Find the Integral sin3xdx Step 1Let . Then , so . Rewrite using and .Tap for more steps...Step . Find .Tap for more steps...Step .Step is constant with respect to , the derivative of with respect to is .Step using the Power Rule which states that is where .Step by .Step the problem using and .Step 2Combine and .Step 3Since is constant with respect to , move out of the 4The integral of with respect to is .Step for more steps...Step and .Step 6Replace all occurrences of with .Step 7Reorder terms.

Theanswer is =-1/5cos^5x+2/3cos^3x-cosx+C We need sin^2x+cos^2x=1 The integral is intsin^5dx=int(1-cos^2x)^2sinxdx Perform the substitution u=cosx, =>, du=-sinxdx Therefore, intsin^5dx=-int(1-u^2)^2du =-int(1-2u^2+u^4)du =-intu^4du+2intu^2du-intdu =-u^5/5+2u^3/3-u =-1/5cos^5x+2/3cos^3x-cosx+C

\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} \nthroot[\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 fx \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 \nu \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} = \ne \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 \notin \exist \mathbb{R} \mathbb{C} \mathbb{N} \mathbb{Z} \emptyset \vee \wedge \neg \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{Radianas} \mathrm{Graus} \square! % \mathrm{limpar} \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{=} + Inscreva-se para verificar sua resposta Fazer upgrade Faça login para salvar notas Iniciar sessão Mostrar passos Reta numérica Exemplos x^{2}-x-6=0 -x+3\gt 2x+1 reta\1,\2,\3,\1 fx=x^3 provar\\tan^2x-\sin^2x=\tan^2x\sin^2x \frac{d}{dx}\frac{3x+9}{2-x} \sin^2\theta' \sin120 \lim _{x\to 0}x\ln x \int e^x\cos xdx \int_{0}^{\pi}\sinxdx \sum_{n=0}^{\infty}\frac{3}{2^n} Mostrar mais Descrição Resolver problemas algébricos, trigonométricos e de cálculo passo a passo step-by-step integral sin^5x pt Postagens de blog relacionadas ao Symbolab Practice Makes Perfect Learning math takes practice, lots of practice. Just like running, it takes practice and dedication. If you want... Read More Digite um problema Salve no caderno! Iniciar sessão
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x²xy+xz=7 y²+yz+yx=13 z²+zx+zy=16 Jika xyz=M/2. Tentukan M. 27. 0.0. Jawaban terverifikasi. Sebuah kolam renang berisi penuh dengan air. Ukuran kolam renang tersebut adalah sebagai berikut: panjang =50 m, lebar =20 meter dân kedalaman =3 meter. Akibat adanya penguapan, kedalaman air berkurang meniadi 2,98m.

$\begingroup$ First off, not going to lie, this is for an assignment. Basically, we're given the integral $$\int \sin^5x\,dx$$ and rewritten form of $$\int [A \sinx + B \sin x \cos^2 x+C\sinx\cos^4x]\,dx$$ using certain trigonometric Identities. We're required to find the values of $A$, $B$ and $C$. Now for the life of me I can't find a set of transformations that will give me that transformation. The power reducing formula gets me to $$\int 5/8\sin X - 5/16\sin3X + 1/16\sin5X $$ and then I can use the multiple angles identity on $\sin3x$ and $\sin5x$, and then I use the power Identities again on the resultant and I just seem to keep going in circles, unable to get the transformation asked for and answer the question. Please send help! egreg235k18 gold badges137 silver badges316 bronze badges asked Sep 23, 2016 at 951 $\endgroup$ 0 $\begingroup$ This is easy. Notice that $$\sin^5 x = \sin x \sin^4 x = \sin x 1- \cos^2 x^2 = \sin x 1 - 2 \cos ^2 x + \cos^4 x ,$$ so $A = 1, \ B = -2, \ C = 1$. Integration, then, is easy, because $$\int \sin x \cos^n x \ \Bbb d x = - \int \cos x' \cos^n x \ \Bbb d x = \frac {\cos^{n+1} x} {n + 1} .$$ answered Sep 23, 2016 at 959 Alex gold badges47 silver badges87 bronze badges $\endgroup$ 2 $\begingroup$Hint You want to find values for $A,B$ and $C$ such that, for all $x$, we have that $$\sin^5x=A\sin x+B\sin x\cos^2x+C\sin x\cos^4x.$$ So try to plug there some specific values, such as $x=\tfrac\pi2$, to solve for $A,B$ and $C$. answered Sep 23, 2016 at 955 WorkaholicWorkaholic6,6332 gold badges22 silver badges57 bronze badges $\endgroup$ You must log in to answer this question. Not the answer you're looking for? Browse other questions tagged .

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integral sin pangkat 5 x dx