Integrate 1 + sinx/cos x

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steve2005

I'm quite stuck on this question: Integrate 1 + sinx/cosx by substitution.
The book also says let u=sin x

Now I was thinking of changing it into sec x + tan x but that didn't get me far

Should I bring cosx to the top?

Sorry btw, I can't use latex so please bear with me!

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Reply 41

raheem94

Original post by Farhan.Hanif93

An alternative approach would be to notice that

\dfrac{1+\sin x}{\cos x} \equiv \dfrac{1+\cos (\frac{\pi}{2} -x)}{\sin (\frac{\pi}{2} - x)} \equiv \dfrac{2\cos^2 (\frac{\pi}{4} - \frac{x}{2})}{2\sin (\frac{\pi}{4} - \frac{x}{2})\cos (\frac{\pi}{4} - \frac{x}{2})}

, which after a bit of cancelling leads to logs directly.

EDIT: just realised the OP was asked for a specific sub, so this can be ignored.

By the way, how do you so quickly think about alternative techniques?

Reply 42

This Honest

Original post by steve2005

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Sorry for the confusion, it's B.

It's sorted though. :smile:

Reply 43

raheem94

Original post by steve2005

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The question is B

Reply 44

f1mad

Original post by raheem94

So we were solving the wrong question. The question was different, right?

Yes, as I pointed out you may have mis-interpreted.

Reply 45

raheem94

Original post by f1mad

Yes, as I pointed out you may have mis-interpreted.

Thanks, but i have understood it before that we were solving the wrong question.

Reply 46

f1mad

Original post by raheem94

Thanks, but i have understood it before that we were solving the wrong question.

No, my post quite clearly indicates I was referring to the question OP had vaguely posted (since the answer is correct).

Reply 47

raheem94

Original post by f1mad

No, my post quite clearly indicates I was referring to the question OP had vaguely posted (since the answer is correct).

Yes, i know OP's question was ambiguous, and we all misunderstood it, nuodai cleared the doubt.

Reply 48

Farhan.Hanif93

Original post by raheem94

By the way, how do you so quickly think about alternative techniques?

Not sure how to answer this. It just stuck out to me as a good approach.

Reply 49

Dirac Spinor

Original post by Farhan.Hanif93

An alternative approach would be to notice that

\dfrac{1+\sin x}{\cos x} \equiv \dfrac{1+\cos (\frac{\pi}{2} -x)}{\sin (\frac{\pi}{2} - x)} \equiv \dfrac{2\cos^2 (\frac{\pi}{4} - \frac{x}{2})}{2\sin (\frac{\pi}{4} - \frac{x}{2})\cos (\frac{\pi}{4} - \frac{x}{2})}

, which after a bit of cancelling leads to logs directly.

That's a cool way but I think your third step took it a bit far. Surely it'd be better to do:

\dfrac{1+\cos (\frac{\pi}{2} -x)}{\sin (\frac{\pi}{2} - x)} \equiv \frac{1-cos(x-\frac{\pi}{2})}{sin(x-\frac{\pi}{2})} \equiv Tan(x/2-\frac{\pi}{4})

Which is standard though I doubt many people know the tan half angle formula.
EDIT: SIGN ERROR!

(edited 12 years ago)

Reply 50

Farhan.Hanif93

Original post by ben-smith

That's a cool way but I think your third step took it a bit far. Surely it'd be better to do:

\dfrac{1+\cos (\frac{\pi}{2} -x)}{\sin (\frac{\pi}{2} - x)} \equiv \frac{1-cos(x-\frac{\pi}{2})}{sin(x-\frac{\pi}{2})} \equiv Tan(x/2-\frac{\pi}{4})

Which is standard though I doubt many people know the tan half angle formula.

I think you might have got your identities wrong here. cos is even and sin is odd so you would still have 1+cos(x-pi/2) for the numerator.

Reply 51

Goku101

Go on to Wolfram alpha and type in integrate "insert question etc" by substitution on search or whatever and then it will show you a complete step by step method.

Someone please tell me why there are some retards negging me!!!!!