Question

this question is integral ans me
${sin}^{4} x$
plz ans me​

Answer 1

qnswer refers to attachment

hope its helpful to you

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Answer 2

Answer:

$\displaystyle \sf \: \int \: {sin}^{4} x \: dx \\ = \sf \: \int \:( { {sin}^{2} \: x})^{2} \: dx \\ \sf \: = \frac{1}{4} \int \: { {(2sin}^{2} \: x})^{2} \: dx \\ \sf \: = \frac{1}{4} \int \: {(1 - cos \: 2x)}^{2} \: dx \\ \sf \: = \frac{1}{4} \int \: (1 - 2cos \: 2x + {cos}^{2} \: x) \: dx \\ \sf \: = \frac{1}{4} \int \: (1 - 2cos \: 2x + \frac{1 + cos \: 4x}{2} ) \: dx \\ \sf \: = \frac{1}{4} \int \: ( \frac{3}{2} - 2cos \: 2x + \frac{1}{ 2} \times cos \: 4x) \: dx \\ \sf \: = \frac{1}{4} \int \: \frac{3}{2} \: dx - \frac{1}{4} \sf \: \int \: 2cos \: 2x \: dx+ \frac{1}{4} \sf \: \int \: \frac{1}{2} \times cos \: 4x \: dx \\ \sf \: = \frac{1}{4} ( \frac{3x}{2} - sin \: 2x + \frac{1}{8} \times sin \: 4x) + c \\ \\ \sf \: where \: \: \: c = \: integral \: constant$