Question

integral of sin^6x dx​

Answer 1

$\large{ \underline{ \underline{ \bigstar{ \: \: \: \: \: \: \pmb{ \sf{Solution \: : }}}}}}$

$\: \: \: \: \: \: \: \: \int {\sin}^{6} (x) dx$

$\dashrightarrow \int { \bigg(1 - { \cos }^{2} (x) \bigg)}^{3} dx$

$\dashrightarrow \int \bigg(1 + {\cos}^{3} (x) - 3 \cos(x) + 3 {\cos}^{2} (x) \bigg) dx$

$\dashrightarrow \int \: dx - \int \frac{1}{4} \bigg( \cos(3x) + 3 \cos(x) \bigg)dx - 3 \int \cos(x) dx + \frac{3}{2} \bigg( \int \cos(2x) dx + \int \: dx \bigg)$

$\dashrightarrow \:x - \frac{1}{4}\sin(3x) - {\frac{3}{4} \sin(x) } - {3 \sin(x) } + \frac{3}{2} \sin(2x) + \frac{3}{2} x + c \: \tt \: \: \: \: \: \: \: \big[ c \: \: is \: \: integral \: \: constant\big]$

$\dashrightarrow { \blue{\frak{ \frac{5}{2} x + \frac{1}{4} \sin(3x) + \frac{3}{2} \sin(2x)-\frac{15}{4}\sin(x)+c }}} \: \: \: \: \: \: \bf ans.$

$\large{ \underline{ \underline{ \bigstar{ \: \: \: \: \: \: \pmb{ \sf{Used \: \: Concepts \: : }}}}}}$

$\tt \rightsquigarrow \: \: \: \cos(3x) = 4 { \cos}^{3} (x) - 3 \cos(x)$

$\tt \rightsquigarrow \: \: \: \cos(2x) = 2{ \cos }^{2} (x) - 1$

$\\ \\ \\ \\ \\ \mathcal{\small \colorbox{aqua}{@MATHOLOGER}}$

Answer 2

Answer:

your answer attached in the photo