Question

integration Tanx/1-sinX​

Answer 1

Multiply denominator and numerator by (1+sinx). The integrand (I) becomes:

$\frac{\tan x(1+\sin x)}{1-\sin^2 x}\\</p><p>=\frac{\sin x}{\cos^3 x }+ \frac{\sin^2 x }{\cos^3 x }\\</p><p>\text{The integral now becomes:}\\</p><p>=\int {\frac{\sin x}{\cos^3 x }}+ \int {\tan^2 x \sec x }\\</p><p>\underline {Solving:}\: \int {\tan^2 x \sec x}\\</p><p>=\int {(\sec x \tan x) \tan x} \: \mathrm{Use \: by \: parts}\\</p><p>\underline {Solving:} \: \int {\frac{sinx}{cos x}}\\</p><p>\implies Put \: u=\cos^3 x\\</p><p>$

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