How do you find the integral of $\int [{sin}^{2} (π x) {cos}^{5} (π x)] d x$ $int [sin^2(pi x) cos^5(pi x)]dx$ ?

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How do you find the integral of $\int [{sin}^{2} (π x) {cos}^{5} (π x)] d x$ $int [sin^2(pi x) cos^5(pi x)]dx$ ?

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Answer 1 · 2015-10-11T02:36:50

This belongs in your mathematical toolbox: $\int {sin}^{m} u {cos}^{n} u d u$ $int sin^m u cos^n u du$ with at least one of $m, n$ $m, n$ odd.

Explanation:

$\int {sin}^{m} u {cos}^{n} u d u$ $int sin^m u cos^n u du$ with at least one of $m, n$ $m, n$ odd.
Integrate by substitution. Do this by pulling off one from the odd power, then convert the remaining even power to the other function. Integrate the resulting polynomial in $sin u$ $sinu$ or $cos u$ $cosu$ term by term.

$I = \int [{sin}^{2} (π x) {cos}^{5} (π x)] d x = \int [{sin}^{2} (π x) {cos}^{4} (π x)] cos (π x) d x$ $I = int [sin^2(pi x) cos^5(pi x)]dx = int [sin^2(pi x) cos^4(pi x)]cos(pix)dx$

${cos}^{4} (π x) = {({cos}^{2} π x)}^{2} = {(1 - {sin}^{2} π x)}^{2} = 1 - 2 {sin}^{2} π x + {sin}^{4} π x$ $cos^4(pi x) = (cos^2 pix)^2 = (1-sin^2 pix)^2 = 1-2sin^2 pix+sin^4 pix$

$I = \int [{sin}^{2} (π x) (1 - 2 {sin}^{2} π x + {sin}^{4} π x)] cos (π x) d x$ $I = int [sin^2(pi x) (1-2sin^2 pix+sin^4 pix)]cos(pix)dx$

$= \int [{sin}^{2} π x - 2 {sin}^{4} π x + {sin}^{6} π x] cos (π x) d x$ $= int [sin^2pix -2sin^4 pix+sin^6 pix]cos(pix)dx$

$= \frac{1}{π} \int (u^{2} - 2 u^{4} + u^{6}) d u$ $= 1/pi int (u^2-2u^4+u^6)du$

$= \frac{1}{π} [\frac{{sin}^{3} π x}{3} - \frac{2 {sin}^{5} π x}{5} + \frac{{sin}^{7} π x}{7}] + C$ $= 1/pi [(sin^3 pix)/3-(2sin^5 pix)/5 + (sin^7 pix)/7] +C$

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How do you find the integral of $\int [{sin}^{2} (π x) {cos}^{5} (π x)] d x$ $int [sin^2(pi x) cos^5(pi x)]dx$ ?

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How do you find the integral of ∫[sin2(πx)cos5(πx)]dxint [sin^2(pi x) cos^5(pi x)]dx?

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How do you find the integral of $\int [{sin}^{2} (π x) {cos}^{5} (π x)] d x$ $int [sin^2(pi x) cos^5(pi x)]dx$ ?