Question

How to integrate dx/1-3sinx

Answer 1

1/2√2 log|(tan(x/2) - 3 - 2√2)/(tan(x/2) - 3 + 2√2)| + C

we have to find Integration of 1/(1 - 3sinx)

using , sin2θ= (2tanθ)/(1 + tan²θ)

so, sinx = 2tan(x/2)/{1 + tan²(x/2)}

now, dx/(1 - 3sinx) = dx/[1 - 6tan(x/2)/{1 + tan²(x/2)}]

= {1 + tan²(x/2)}dx/{1 + tan²(x/2) - 6tan(x/2)}

= sec²(x/2)dx/{1 + tan²(x/2) - 6tan(x/2)}

now let tan(x/2) = p

differentiating both sides,

sec²(x/2) dx = 2dp

so, 2dp/{1 + p² - 6p}

= 2dp/{(p - 3)² - (2√2)²}

from application of Integration,

dx/(x² - a²) = 1/2a ln|(x - a)(x + a)| + C

so, dp/{(p - 3)² - (2√2)²} = 1/4√2 log|(p - 3 - 2√2)/(p - 3 + 2√2)|

so, dx/(1 - 3x) = 2 × 1/4√2 log|(tan(x/2) - 3 - 2√2)/(tan(x/2) - 3 + 2√2)| + C

=1/2√2 log|(tan(x/2) - 3 - 2√2)/(tan(x/2) - 3 + 2√2)| + C

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