Dear student, The solution is: ∫1sin x +tanxdxLet u = tan (x2)Then dx=21+u2du , sin x = 2u1+u2, tan x = 2u1-u2∫1sin x +tanxdx =∫12u1+u2+2u1-u221+u2du=∫1-u22udu=12∫1-u2udu=12∫(1u-u )du=12(∫1u du -∫udu)=12ln(u)- u22Now putting back u =tan ( x2) we get∫1sin x +tanxdx =12lntan(x2))-tan 2(x2)2+C Regards

Please solve Q. No.25

Dear student,
The solution is:

\int \frac{1}{\sin x + \tan x} d x L e t u = \tan (\frac{x}{2}) T h e n d x = \frac{2}{1 + u^{2}} d u, \sin x = \frac{2 u}{1 + u^{2}}, \tan x = \frac{2 u}{1 - u^{2}} \int \frac{1}{\sin x + \tan x} d x = \int \frac{1}{\frac{2 u}{1 + u^{2}} + \frac{2 u}{1 - u^{2}}} \frac{2}{1 + u^{2}} d u = \int \frac{1 - u^{2}}{2 u} d u = \frac{1}{2} \int \frac{1 - u^{2}}{u} d u = \frac{1}{2} \int (\frac{1}{u} - u) d u = \frac{1}{2} (\int \frac{1}{u} d u - \int u d u) = \frac{1}{2} \{\ln (u) - \frac{u^{2}}{2}\} N o w p u t t i n g b a c k u = \tan (\frac{x}{2}) w e g e t \int \frac{1}{\sin x + \tan x} d x = \frac{1}{2} \{\ln \tan (\frac{x}{2})) - \frac{\tan^{2} (\frac{x}{2})}{2}\} + C

Regards

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