Question

my question is: Integral ( 1 / (1 - sec x) )

Answer 1

what if you add zero to the top? like say (sec-sec) ?

then split the fraction

Answer 2

Umm make it easy for me,, if you can give me the full answer will be better.

Answer 3

\[\int \frac{1}{1-sec}\]
\[\int \frac{1+sec-sec}{1-sec}\]
\[\int \frac{1-sec}{1-sec}+\frac{sec}{1-sec}\]
\[\int 1+\frac{sec}{1-sec}\]

or we might be able to say sec(x) = u to begin with and construct an equivalent integral from us

Answer 4

or if we multiply by the conjugate that might get it into something doable

Answer 5

\[\int \frac{1}{1-sec}\]
\[\int \frac{1}{1-sec}*\frac{1+sec}{1+sec}\]
\[\int \frac{1+sec}{1+sec^2}\]
but maybe not lol

Answer 6

my teacher use this way: put u = some thing... then complate

Answer 7

u = tan(x/2) perhaps?

Answer 8

i never do well with us and trigs combined ... gives me a headache

Answer 9

\[\frac{\cos{x}}{\cos{x} - 1}\]
\[ \frac{\cos ^{2}\frac{x}{2} - \sin^{2}\frac{x}{2}}{\cos ^{2}\frac{x}{2} - \sin^{2}\frac{x}{2} -1}\]

Answer 10

I have got a headache from this question

Answer 11

i tried to follow wolframs steps and got even more discombobulated

Answer 12

\[\frac{1 - \tan^{2} \frac{x}{2}}{1 - \tan^{2}\frac{x}{2} - \sec^{2}\frac{x}{2}}\]

Answer 13

ishaan94 there is no cos in the question!! give me more explan.

Answer 14

\[\tan \frac{x}{2} = u\]
\[ \frac12 \sec^{2}\frac{x}{2}= \frac{du}{dx}\]
\[dx = \frac{2\times du}{\sec^{2}\frac{x}{2}}\]
\[\frac{1 - u^2}{\left(1 - u^2 -(1 + u^2)\right)\left(1 + u^2\right)}2 du\]
\[\frac{1 - u^2}{-2u^2(1 + u^2)}2du\]
\[-\frac{1-u^2}{u^2(1 + u^2)}du\]

Answer 15

Use partial fraction or something else...I am outta here Good Luck

Answer 16

OK thanks.. I will try.