what is the integral of sin x You can also show that this is equivalent to the other two answers using the identity cos 2x cos2 x sin2 x Depending on the route you take valid results include sin 2 x 2 C cos 2 x 2 C 1 4cos 2x C There are a variety of methods we can take Substitution with sine Let u sin x This implies that du cos x dx
What is the integral of int sin 5 x dx Calculus Introduction to Integration Integrals of Trigonometric Functions 1 Answer Show the steps below Substitute u 2 x dx 2 du sin x2 dx 2 sin u2 2 du These is special integral Fresnel integral S u Plug in solved integrals 2 sin u2 2 du S u 2
what is the integral of sin x
what is the integral of sin x
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The Integral Of Sin Squared X YouTube
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What Is The Integral Of Sin X Dx From X 0 To X Pi Week 12
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Answer link intarcsin x dx xarcsin x sqrt 1 x 2 C We will proceed by using integration by substitution and integration by parts Substitution Let t arcsin x x sin t and dx cos t dt Then substituting we have intarcsin x dx inttcos t dt Integration by Parts Let u t and dv cos t dt Then du dt and v sin t By Using the definition of the integral and the fact that sinx is an odd function from 0 to 2pi with equal area under the curve at 0 pi and above the curve at pi 2pi the integral is 0 This holds true for any time sinx is evaluated with an integral across a domain where it is symmetrically above and below the x axis int 0 2pi sinxdx cosx 0 2pi cos2pi
Putting it all together we get our final result sin3 x dx sin x dx sin x cos2 x dx cos x 1 3cos3 x C Answer link intsin 3 x dx 1 3cos 3 x cos x C intsin 3 x dx intsin x 1 cos 2 x dx intsin x dx intsin x cos 2 x dx For the first integral intsin x dx cos x C For the second integral using What is the integral of sin 6 x Calculus Basic Differentiation Rules Chain Rule 1 Answer maganbhai P
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Explanation We should try to use substitution by setting u cosx so du sinxdx This gives us the integral sinx cos2x dx sinx cos2x dx 1 u2 du u 2du From here use the rule undu un 1 n 1 C Thus Explanation Use integration by parts which takes the form udv uv vdu For udv x2sin x dx we let u x2 du dx 2x du 2xdx dv sin x dx dv sin x dx v cos x Thus substituting these into the integration by parts formula we see that x2sin x dx x2cos x 2xcos x dx
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