Answer(1 of 4): I think you mean ∫[x+(1/x)]³ dx=∫f(x)dx ,for which you have no choice but to expand the INTEGRAND in the smart way ① f(x)=[x+(1/x)]³ Step-by-Step Examples Calculus Integral Calculator Step 1 Enter the function you want to integrate into the editor. The Integral Calculator solves an indefinite integral of a function. You can also get a better visual and understanding of the function and area under the curve using our graphing tool. Integration by parts formula ?udv=uv-?vdu Step 2 Click the blue arrow to submit. Choose "Evaluate the Integral" from the topic selector and click to see the result!

ContohSoal Integral Substitusi. Berikut ini adalah contoh soal integral substitusi aljabar beserta dengan pembahasannya, simak baik-baik ya! Tentukanlah integral dari. Nah untuk menjawab soal integral di atas, kita ambil pemisalan. Biasanya yang di dalam tanda kurung atau di dalam tanda akar atau yang pangkatnya paling besar.

$\begingroup$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 . Tentukanintegral fungsi aljabar (x pangkat 2 - 5x) dx. Question from @Fiolet17 - Sekolah Menengah Atas - Matematika. Metode persial pada soal berikut! integral 2x sin (3x -1) dx Answer. Fiolet17 April 2019 | 0 Replies . Tentukan nilai integral berikut! integral (10x pangkat 4 + 6x pangkat 3 + 2) dx The answer is =-1/5cos^5x+2/3cos^3x-cosx+C Explanation We need sin^2x+cos^2x=1 The integral is intsin^5dx=int1-cos^2x^2sinxdx Perform the substitution u=cosx, =>, du=-sinxdx Therefore, intsin^5dx=-int1-u^2^2du =-int1-2u^2+u^4du =-intu^4du+2intu^2du-intdu =-u^5/5+2u^3/3-u =-1/5cos^5x+2/3cos^3x-cosx+C Rozwiązujzadania matematyczne, korzystając z naszej bezpłatnej aplikacji, która wyświetla rozwiązania krok po kroku. Obsługuje ona zadania z podstaw matematyki, algebry, trygonometrii, rachunku różniczkowego i innych dziedzin.
\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
Contoh4. ∫ (x2+ 1)5.2x dx = (x2+ 1)6/6 + C. (Disini kita menerapkan Aturan Pangkat yang Diperumum dengan g(x) = x2 + 1, g'(x) = 2x.) Contoh 5. Jika g(x) = sin x, maka g'(x) = cos x. Jadi, menurut Aturan Pangkat yang Diperumum, diperoleh ∫ sin dx = (sin x)2/2 + C. Latihan. Tentukan integral tak tentu di bawah ini. 1. ∫(x2+ x-2
\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] â–­\\longdivision{â–­} \times \twostack{â–­}{â–­} + \twostack{â–­}{â–­} - \twostack{â–­}{â–­} \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{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{=} + Subscribe to verify your answer Subscribe Sign in to save notes Sign in Show Steps Number Line Examples x^{2}-x-6=0 -x+3\gt 2x+1 line\1,\2,\3,\1 fx=x^3 prove\\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} Show More Description Solve problems from Pre Algebra to Calculus step-by-step step-by-step \int \sin5xdx en Related Symbolab blog posts Practice Makes Perfect Learning math takes practice, lots of practice. Just like running, it takes practice and dedication. If you want... Read More Enter a problem Save to Notebook! Sign in
Explanation Let I = ∫ x2 +3x− 43x +2 dx = ∫ (x−1)(x +4)3x +2 dx You actually didn't use the chain rule correctly: To use the chain rule, you need to also multiply 2(1−x) by dxd (1−x) = −1. In other words, we have f (x) = (1− x)2 f ′(x) = 2(1− x)⋅ dxd (1− x) = −2(1− x) = 2(x −1) How to formalize a variable The equation can be written as On separating the integrals As we know, dcos x = - sin x dx Therefore, put cos x = t and dt = - sin x dx in above TheIntegral Calculator lets you calculate integrals and antiderivatives of functions online — for free! Our calculator allows you to check your solutions to calculus exercises. It helps you practice by showing you the full working (step by step integration). All common integration techniques and even special functions are supported. $\begingroup$ 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 .
5atau 5 dx dy x dy dx x . Contoh 6: Luas daerah yang dibatasi oleh kurva y = f(x) dengan f(x) > 0, sumbu x, dan dua garis tegak, yang pertama tetap dan yang kedua variabel, diketahui sama dengan tiga kali panjang kurva tersebut diantara kedua buah garis tegak tersebut. Tentukan persamaan diferensial dari kurva f ! Jawab:
\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 Daftarintegral dari fungsi trigonometri Daftar integral trigonometri (antiderivatif: integral tak tentu) dari fungsi trigonometri. Untuk antiderivatif yang melibatkan baik fungsi eksponensial dan trigonometri, lihat Daftar integral dari fungsi eksponensial. = ∫ (3/8 + 1/2 cos 2x + 1/8 cos 4x) dx = 3/8 x + 1/4 sin 2x + 1/32 sin 4x + c. 5.

This integral is mostly about clever rewriting of your functions. As a rule of thumb, if the power is even, we use the double angle formula. The double angle formula says sin^2theta=1/21-cos2theta If we split up our integral like this, int\ sin^2x*sin^2x\ dx We can use the double angle formula twice int\ 1/21-cos2x*1/21-cos2x\ dx Both parts are the same, so we can just put it as a square int\ 1/21-cos2x^2\ dx Expanding, we get int\ 1/41-2cos2x+cos^22x\ dx We can then use the other double angle formula cos^2theta=1/21+cos2theta to rewrite the last term as follows 1/4int\ 1-2cos2x+1/21+cos4x\ dx= =1/4int\ 1\ dx-int\ 2cos2x\ dx+1/2int\ 1+cos4x\ dx= =1/4x-int\ 2cos2x\ dx+1/2x+int\ cos4x\ dx I will call the left integral in the parenthesis Integral 1, and the right on Integral 2. Integral 1 int\ 2cos2x\ dx Looking at the integral, we have the derivative of the inside, 2 outside of the function, and this should immediately ring a bell that you should use u-substitution. If we let u=2x, the derivative becomes 2, so we divide through by 2 to integrate with respect to u int\ cancel2cosu/cancel2\ du int\ cosu\ du=sinu=sin2x Integral 2 int\ cos4x\ dx It's not as obvious here, but we can also use u-substitution here. We can let u=4x, and the derivative will be 4 1/4int\ cosu\ dx=1/4sinu=1/4sin4x Completing the original integral Now that we know Integral 1 and Integral 2, we can plug them back into our original expression to get the final answer 1/4x-sin2x+1/2x+1/4sin4x+C= =1/4x-sin2x+1/2x+1/8sin4x+C= =1/4x-1/4sin2x+1/8x+1/32sin4x+C= =3/8x-1/4sin2x+1/32sin4x+C

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    • integral sin pangkat 5 x dx