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adamk · Answer

Let's hear it.

nothingwasthesame · Answer

ok

anonymous · Answer

really stuck on
$$\int\limits \frac{1}{1-sinx}dx$$
any pointers appreciated while i do the next ques....

ribhu · Answer

first rationalize the denominator.

ribhu · Answer

what do u get?

anonymous · Answer

ok

anonymous · Answer

let me try

solomonzelman · Answer

$\large\color{slate}{\displaystyle\int\limits_{~}^{~}\frac{1}{1-\sin x}dx}$

$\large\color{slate}{\displaystyle\int\limits_{~}^{~}\frac{1+\sin x}{1-\sin^2x}dx}$

$\large\color{slate}{\displaystyle\int\limits_{~}^{~}\frac{1+\sin x}{\cos^2x}dx}$

$\large\color{slate}{\displaystyle\int\limits_{~}^{~}\frac{1}{\cos^2x}+\frac{\sin x}{\cos^2x}dx}$      I think this is a direction to take.

anonymous · Answer

$$\int\limits \frac{1+sinx}{\cos^2x}dx$$
$$\int\limits(\sec^2x+secxtanx)dx=\int\limits \sec^2xdx+\int\limits secxtanxdx$$

anonymous · Answer

$$tanx+secx+C$$

solomonzelman · Answer

yes, and those are recognizable derivatives

solomonzelman · Answer

Yes.

anonymous · Answer

aww man didn't think about rationalizing ahh..

solomonzelman · Answer

your answer is right. but put \ in front of a trigonometric function for a better code.

solomonzelman · Answer

` \sin(x) ` like this for example

ribhu · Answer

yeah did u develop the insight for these kind of problems?

anonymous · Answer

thanks :D

solomonzelman · Answer

yes, rationalizing is a good tool to break it up when dealing with trig functions.

ribhu · Answer

@Nishant_Garg ideas will float once you practice.

anonymous · Answer

sorry I lost connection, i'll close the question now