Let ∫dx2sinx+secx=alogetanb+x2−secc−x+d, where a, b, c, d are constants and b,c∈0,π2, then

∫dx2sinx+secx=∫cosx dx2sinxcosx+1=12∫(cosx+sinx)+(cosx−sinx)(cosx+sinx)2dx=12∫dxcosx+sinx+12∫cosx−sinx(cosx+sinx)2dx=122∫secx−π4dx−121cosx+sinx+C1=122lntanπ8+x2−122secπ4−x+C

Let ∫dx2sinx+secx=alogetanb+x2−secc−x+d, where a, b, c, d are constants and b,c∈0,π2, then

∫dx2sinx+secx=∫cosx dx2sinxcosx+1=12∫(cosx+sinx)+(cosx−sinx)(cosx+sinx)2dx=12∫dxcosx+sinx+12∫cosx−sinx(cosx+sinx)2dx=122∫secx−π4dx−121cosx+sinx+C1=122lntanπ8+x2−122secπ4−x+C

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Let $\int \frac{dx}{2 sinx + secx} = a [\log_{e} |\tan (b + \frac{x}{2})| - \sec (c - x)] + d,$ where a, b, c, d are constants and $b, c \in (0, \frac{π}{2}),$ then

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$a^{2} - \frac{b}{π} = \frac{1}{8}$

$b + c = \frac{3 π}{8}$

$a^{2} + \frac{b}{π} = \frac{5}{8}$

$b - c = \frac{π}{8}$

answer is B.

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Detailed Solution

$\int \frac{dx}{2 sinx + secx} = \int \frac{cosx dx}{2 sinxcosx + 1} = \frac{1}{2} \int \frac{(cosx + sinx) + (cosx - sinx)}{{(cosx + sinx)}^{2}} dx$
$\begin{array}{l} = \frac{1}{2} \int \frac{dx}{cosx + sinx} + \frac{1}{2} \int \frac{cosx - sinx}{{(cosx + sinx)}^{2}} dx \\ = \frac{1}{2 \sqrt{2}} \int \sec (x - \frac{π}{4}) dx - \frac{1}{2} \frac{1}{cosx + sinx} + C_{1} \end{array}$
$= \frac{1}{2 \sqrt{2}} \ln |\tan (\frac{π}{8} + \frac{x}{2})| - \frac{1}{2 \sqrt{2}} \sec (\frac{π}{4} - x) + C$