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Answer 1

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student-name Ashik Ak asked in Math

integrate sin^4x dx

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student-name Vikramjeet Singh answered this

6 helpful votes in Math, Class XII-Science

sin^4(x)

=>[sin^2(x)]^2

=>[(1 -cos(2x)/2)]^2

=>1/4[1 + cos^2(2x) - 2cos(2x)]

=>1/4[1 - 2cos(2x) + (1 + cos(4x)/2 ]

=>(1/4) - (1/2)cos(2x) + (1/8) + (1/8)cos(4x)

=>(3/8) - (1/2)cos(2x) + (1/8)cos(4x)

so ∫sin^4(x)dx = 3/8∫dx - 1/2∫cos(2x) dx + 1/8∫cos(4x) dx

∫sin^4(x)dx = (3/8)x - (1/2)sin(2x)*(1/2) + (1/8)sin(4x)*(1/4) + c

∫sin^4(x)dx = (3/8)x - (1/4)sin(2x) + (1/32)sin(4x) + c

Answer 2

$\frac{3x}{8} - \frac{3 \sin(x) \cos(x) }{8} - \frac{ { \sin(x) }^{3} \cos(x) }{4} + c$