How do I evaluate the indefinite integral $\int {sin}^{6} (x) \cdot {cos}^{3} (x) d x$ $intsin^6(x)*cos^3(x)dx$ ?

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Answer 1 · 2014-09-21T19:27:45

By substitution,

$\int {sin}^{6} x {cos}^{3} x d x = \frac{{sin}^{7} x}{7} - \frac{{sin}^{9} x}{9} + C$ $int sin^6x cos^3x dx={sin^7x}/7-{sin^9x}/9+C$

Let us look at some details.

$\int {sin}^{6} x {cos}^{3} x d x$ $int sin^6x cos^3x dx$

by pulling out $cos x$ $cosx$ ,

$= \int {sin}^{6} x {cos}^{2} x \cdot cos x d x$ $=int sin^6x cos^2x cdot cosx dx$

by the trig identity ${cos}^{2} x = 1 - {sin}^{2} x$ $cos^2x=1-sin^2x$ ,

$= \int {sin}^{6} x (1 - {sin}^{2} x) \cdot cos x d x$ $=int sin^6x(1-sin^2x)cdot cosx dx$

by the substitution $u = sin x$ $u=sinx$ . $\Rightarrow d u = cos x d x$ $Rightarrow du=cosx dx$ ,

$= \int u^{6} (1 - u^{2}) d u = \int u^{6} - u^{8} d u$ $=int u^6(1-u^2)du=int u^6-u^8 du$

by Power Rule,

$= \frac{u^{7}}{7} - \frac{u^{9}}{9} + C$ $=u^7/7-u^9/9+C$

by putting $u = sin x$ $u=sinx$ back in,

$= \frac{{sin}^{7} x}{7} - \frac{{sin}^{9} x}{9} + C$ $={sin^7x}/7-{sin^9x}/9+C$

How do I evaluate the indefinite integral ∫sin6(x)⋅cos3(x)dxintsin^6(x)*cos^3(x)dx ?