Question

sin⁶x dx integrate the following​

Answer 1

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Question :

$\displaystyle \spadesuit\ \; \sf \red{\int sin^6x\ dx}$

Solution :

$\rm sin^6x$

$:\implies\ \; \rm (sin^2x)^3$

$\\ :\implies\ \; \rm \left( \dfrac{1-cos\ 2x}{2} \right) ^3$

$\\ :\implies\ \; \rm \dfrac{1}{8} \left( 1-cos^3(2x) - 3\ cos(2x) + 3\ cos^2(2x) \right)$

$\bullet\ \; \pink{\bf cos^2x= \dfrac{1+cos\ (2x)}{2}}$

$\bullet\ \; \green{\bf cos^3x= \dfrac{cos\ (3x)+3\ cos\ x}{4}}$

$\\ :\implies\ \; \rm \dfrac{1}{8} \left( 1- 3\ cos(2x) + \dfrac{3}{2}\ +\dfrac{3}{2}\ cos(4x) - \dfrac{1}{4}\ cos(6x) - \dfrac{3}{4}\ cos(2x) \right)$

$\\ :\implies\ \; \rm \dfrac{1}{8} \left( \dfrac{5}{2} - \dfrac{15}{4}\ cos(2x) +\dfrac{3}{2}\ cos(4x) - \dfrac{1}{4}\ cos(6x) \right)$

So ,

$\displaystyle \sf \red{\int sin^6x\ dx}$

$\\ \displaystyle :\implies\ \; \rm \dfrac{1}{32} \int \left( 10 - 15\ cos(2x) +6\ cos(4x) - cos(6x) \right)\ dx$

$\\ \displaystyle :\implies\ \; \rm \dfrac{1}{32} \left( 10x - \dfrac{15}{2}\ sin(2x) +\dfrac{6}{4}\ sin(4x) - \dfrac{1}{6}\ sin(6x) \right)$

$\\ \displaystyle :\implies\ \; \bf \dfrac{1}{384} \left( 120x - 90\ sin(2x) +18\ sin(4x) - 2\ sin(6x) \right)\ +c$

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