Calculate-B-1-1-sin-x-dx-using-u-tan-x-2-

Question Number 169250 by mathocean1 last updated on 26/Apr/22

C a l c u l a t e B = \int \frac{1}{1 + s i n (x)} d x u s i n g u = t a n (\frac{x}{2})

Commented by infinityaction last updated on 27/Apr/22

I = \int \frac{d x}{{(\sin \frac{x}{2} + \cos \frac{x}{2})}^{2}}

I = \int \frac{\sec^{2} x / 2}{{(1 + \tan x / 2)}^{2}} d x

u = \tan x / 2

d u = \frac{1}{2} \sec^{2} x / 2

I = \int \frac{2 d u}{{(1 + u)}^{2}}

I = - \frac{2}{1 + u} + c

I = - \frac{2}{1 + \tan x / 2} + c

Answered by MikeH last updated on 27/Apr/22

\sin x = \frac{2 t}{1 + t^{2}}

d x = \frac{2 d t}{1 + t^{2}}

\Rightarrow B = \int \frac{1}{(1 + \frac{2 t}{1 + t^{2}})} . \frac{2 d t}{1 + t^{2}} = \int \frac{2 d t}{t^{2} + 2 t + 1}

B = 2 \int \frac{d t}{{(1 + t)}^{2}} = - \frac{2}{1 + t} + k

B = - \frac{2}{1 + \tan (\frac{x}{2})} + k

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