Evaluatethe integral: ∫x sin x cos x dx. asked Jun 26, 2020 in Indefinite Integral by Vikram01 (51.7k points) methods of integration; class-12; 0 votes. 1 answer. Evaluate the integral: ∫x cos 2x dx. asked Jun 26, 2020 in Indefinite Integral by Vikram01 (51.7k points) methods of integration; class-12; 0 votes. 1 answer.
integralx^-4 (x+5)^2 dx adalah Tanya. 11 SMA. Matematika. KALKULUS.
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a ∫ (3x 7 − π) dx = Jawab : = \(\mathrm{\frac{3}{7+1}}\)x 7+1 − πx + C = \(\mathrm{\frac{3}{8}}\)x 8 − πx + C b. ∫ (6x 5 + 2x 3 − x 2) dx
³kfk³ f(x )dx (ii) ³ f (x ) g (x )dx ³ f (x )dx ³ g (x )dx Latihan : Cari integral tak tentu berikut : a. ³x 3 x 1 dx b. ³(y 4y)2 dy c. dx x x 3 3x 2 1 ³ d. ³3sin 2cost dt e. dx x x x n 2c 3n2 ³ Teorema 4 : Substitusi Integral Tak Tentu Misal g adalah fungsi yang dapat diturunkan dan F adalah suatu anti turunan dari f. Jika u g(x) maka
. $\begingroup$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 .
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integral sin pangkat 5 x dx
