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intsin^(5)xcos^(5)xdx

sin5xcos5xdx \int \sin ^{5} x \cos ^{5} x d x

Full solution

Q. sin5xcos5xdx \int \sin ^{5} x \cos ^{5} x d x
  1. Identify Substitution: Identify a substitution that can simplify the integral.\newlineWe can use the substitution u=sin(x)u = \sin(x), which implies du=cos(x)dxdu = \cos(x)dx.
  2. Rewrite in terms of u: Rewrite the integral in terms of u and du.\newlineThe integral becomes u5(1u2)2du\int u^5(1-u^2)^2\,du, since cos2(x)=1sin2(x)\cos^2(x) = 1 - \sin^2(x) and we have cos5(x)=cos(x)(1sin2(x))2\cos^5(x) = \cos(x) \cdot (1 - \sin^2(x))^2.
  3. Expand Integrand: Expand the integrand.\newlineWe need to expand (1u2)2(1-u^2)^2 to integrate term by term. The expansion is (12u2+u4)(1 - 2u^2 + u^4).
  4. Multiply by u5u^5: Multiply the expanded terms by u5u^5.\newlineMultiplying u5u^5 by each term in the expansion gives us u52u7+u9u^5 - 2u^7 + u^9.
  5. Integrate Each Term: Integrate each term separately.\newlineThe integral of u5u^5 is (u6)/6(u^6)/6, the integral of 2u7-2u^7 is (2u8)/8(-2u^8)/8, and the integral of u9u^9 is (u10)/10(u^{10})/10.
  6. Combine and Simplify: Combine the integrated terms and simplify.\newlineCombining the terms, we get (u6)/6(u8)/4+(u10)/10(u^6)/6 - (u^8)/4 + (u^{10})/10.
  7. Substitute back in xx: Substitute back in terms of xx.\newlineSince u=sin(x)u = \sin(x), we substitute back to get sin6(x)6sin8(x)4+sin10(x)10+C\frac{\sin^6(x)}{6} - \frac{\sin^8(x)}{4} + \frac{\sin^{10}(x)}{10} + C, where CC is the constant of integration.

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