Evaluate the following integrals. ∫sin2xdx

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Evaluate the following integrals.  ∫sin2xdx

Eprint · Accepted Answer

Step 1    We must assess the integral ∫sin2xdx    Step 2    ∫sin2xdx    as cos⁡2x=cos2x−sin2x    cos⁡2x=1−2sin2x    sin2x=1−cos⁡2x2    therefore    ∫1−cos⁡2x2dx    ⇒12∫(1−cos⁡2x)dx    ⇒12∫dx−12∫cos⁡2xdx    ⇒12x−12(sin⁡2x2)+C    ⇒12x−14sin⁡2x+C    Therefore    ∫sin2xdx=12x−14sin⁡2x+C

Leory2000 · Accepted Answer

Step 1: Use Pythagorean Identities: sin2x=12−cos⁡2x2.  ∫12−cos⁡2x2dx  Step 2: Use Sum Rule: ∫f(x)+g(x)dx=∫f(x)dx+∫g(x)dx.  ∫12dx−∫cos⁡2x2dx  Step 3: Use this rule: ∫adx=ax+C.  x2−∫cos⁡2x2dx  Step 4: Use Constant Factor Rule: ∫cf(x)dx=c∫f(x)dx.  x2−12∫cos⁡2xdx  Step 5: Use Integration by Substitution on ∫cos⁡2xdx.  Let u=2x, du=2dx, then dx=12du  Step 6: Using u and du above, rewrite ∫cos⁡2xdx.  ∫cos⁡u2du  Step 7: Use Constant Factor Rule: ∫cf(x)dx=c∫f(x)dx.  12∫cos⁡udu  Step 8: Use Trigonometric Integration: the integral of cos⁡u is sin⁡u.  sin⁡u2  Step 9: Substitute u=2x back into the original integral.  sin⁡2x2  Step 10: Rewrite the integral with the completed substitution.  x2−sin⁡2x4  Step 11: Add constant.  x2−sin⁡2x4+C

Evaluate the following integrals. \int \sin^{2}x dx

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