Hard integral - maybe

Without using

\tan (x/2)

substitution, evaluate the integral

\displaystyle \int_{0}^{\frac{1}{4} \pi } \frac{1+\sin x}{1-\sin x} dx

Without using

\tan (x/2)

substitution, evaluate the integral

\displaystyle \int_{0}^{\frac{1}{4} \pi } \frac{1+\sin x}{1-\sin x} dx

Are you asking a question or posing a challenge? In any case multiply the denominator and numerator by

1+\sin x

so the integrand becomes

\bigg(\dfrac{1+\sin x}{\cos x}\bigg)^{2}

. Then use the relation

\dfrac{1+\sin x}{\cos x} = \cot \left(\frac{x}{2} + \frac{\pi}{8}\right)

to evaluate the integral.

Spoiler

(edited 7 years ago)

Ok. A harder one I think.

\displaystyle \int_{b}^{a+2b} \sqrt{ \frac{x-b}{x+a}} \text{ d}x

(edited 7 years ago)

Ok. A harder one I think.

\displaystyle \int_{b}^{a+2b} \sqrt{ \frac{x-b}{x+a}} \text{ d}x

Substitute

u = \sqrt{\frac{x-b}{x+a}}

; it becomes easy partial fractions after that.

Substitute

u = \sqrt{\frac{x-b}{x+a}}

; it becomes easy partial fractions after that.

I used sub

x=a\sinh^2 t + b\cosh^2 t

. I made this question up and didn't see the sun you made. :wink:

Did you get an answer or just spot the substitution without actually doing it?

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