int q (1 Viewer)

azureus88 · Mar 3, 2009

Just wondering if there is an alternative method to t method to solve int[1/(1+sinx)]. I know t method doesnt take long but just wondering if there are alternatives.

Any replies would be appreciated

vds700 · Mar 3, 2009

you could do it this wau i guess
$\int \frac{dx}{1+sinx}\\=\int \frac{1}{1+sinx}.\frac{1-sinx}{1-sinx}\\=\int \frac{(1-sinx)dx}{1-sin^{2}x}\\=\int\frac{(1-sinx)dx}{cos^{2}x}\\=\int (sec^{2}x-\frac{sinx}{cos^{2}x})dx$ now just integrate

azureus88 · Mar 3, 2009

k thanks. wat about int[sinx/(1+sin^2x)]?

Drongoski · Mar 3, 2009

azureus88 said:
k thanks. wat about int[sinx/(1+sin^2x)]?

Int (sin x/(1 + sin 2x))dx = Int[sin x/2cos^2 x)] dx = -(1/2)Int (cos x )^(-2)d(cos x)
= 1/(2cos x) + C

azureus88 · Mar 3, 2009

Drongoski said:
Int (sin x/(1 + sin 2x))dx = Int[sin x/2cos^2 x)] dx = -(1/2)Int (cos x )^(-2)d(cos x)
= 1/(2cos x) + C

The question was actually sin squared x not sin2x. actually dont worry, i think i got it now.

lolokay · Mar 3, 2009

$\int \frac{sin[x]}{1+sin^2 [x]}dx\\ = \int \frac{sin[x]}{2-cos^2 [x]}dx\\ = \frac{\sqrt2}{4}\int \frac{sinx}{\sqrt2-cos[x]}+\frac{sinx}{\sqrt2+cos[x]}dx\\ using \ partial \ fractions\\ = \frac{\sqrt2}{4}(ln|\sqrt2-cos[x]| - ln|\sqrt2+cos[x]|) + C$

is one way to do it

Timothy.Siu · Mar 3, 2009

nice, i shall remember these methods

int q (1 Viewer)

azureus88

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